ENGINEERING 
LIBRARY 


NOTES 


RANKINE'S 
APPLIED  MECHANICS. 


THE  LIBRARY 

OF 

THE  UNIVERSITY 
OF  CALIFORNIA 

LOS  ANGELES 

GIFT  OF 

U.  of  Calif.  . 
Berkeley 


'7- 


XT, 


NOTES 


RANKINGS  APPLIED  MECHANICS. 


GEORGE   I.   ALDEN,  B.  S., 

PROFESSOR   OF    THEORETICAL,   AND   APPLIED  MECHANICS 

IN    THE    WORCESTER    FREE    INSTITUTE, 

WORCESTER,    MASS. 


Third  Edition. 


HARTFORD,    CONN.: 

PRESS  OF  THE  CASE,  LOCKWOOD  &  BRAINARD  COMPANY. 
1890. 


Copyright 
BY  GEORGE  I.  ALDEN. 

1877. 


TA 

35"° 

All* 


INTKODTJOTI 


The  following  pages  are  the  -result  of  putting  in  permanent 
form  some  of  the  matter  which  it  ha.s  been  found  expedient  or 
necessary  to  give  by  dictation  to  students  in  the  Worcester 
Free  Institute,  who  pursue  for  the  first  time,  the  study  of  Ran- 
kine's  Applied  Mechanics.  The  object  in  their  publication  is  not 
to  furnish  a  key,  or  provide  a  substitute  for  diligent  study  and 
careful  thought  on  the  part  of  the  student,  but  rather  to  en- 
courage him  by  giving  such  suggestions,  solutions,  and  refer- 
ences as  experience  has  shown  that  the  average  student  requires; 
thus  economizing  time  in  the  preparation  of  the  lesson,  and  also 
giving  the  instructor  opportunity  to  devote  the  time  spent  in 
the  class  room  to  recitations,  and  to  the  application  of  the  prin- 
ciples and  formulae  of  the  lesson,  to  practical  problems. 

To  what  may  be  strictly  called  notes  on  the  "Applied  Me- 
chanics," I  have  added  a  brief  explanation  and  illustration  of  the 
method  of  producing  the  reciprocal  diagram  of  stresses,  sub- 
stantially taken  from  "  Economics  of  Construction,"  by  R.  E. 
Bow,  C.  E.  Also  a  separate  treatise  on  strength  of  beams,  and 
an  investigation  of  a  particular  problem  relating  to  seven  bar 
parallel  motions,  known  as  "  Peaucillier's  Parallel  Motion." 

This  work  has  been  prepared  from  materials  drawn  from 
various  sources,  especially  from  notes  given  by  Prof.  Eustis,  of 
The  Lawrence  Scientific  School,  to  the  class  of  '68. 

I  have  also  received  assistance  from  George  H.  White,  B.  S., 
a  graduate  of  the  Free  Institute,  and  have  inserted  on  several 
articles  of  the  Applied  Mechanics,  notes  which  are  entirely  his 
own  work.  I  have  endeavored  to  make  proper  reference  to 
works  from  which  quotations  or  extracts  have  been  taken. 

The  blank  pages  at  the  end  are  intended  to  receive  such  sup- 
plementary notes  as  the  instructor  may  find  adapted  to  the 
needs  and  capacity  of  his  class. 

GEORGE  I.  ALDEN. 

WOBCESTEE  FREE  INSTITUTE,  Feb.  1st,  1877. 

NOTE. —  Some  errors  found  in  the  earlier  editions  of  "Rankine's 
Applied  Mechanics  "  have  been  corrected  in  later  editions,  issued  since 
these  notes  were  prepared.  References  in  the  notes,  to  such  errors, 
are  allowed  to  remain  for  the  benefit,  of  those  who  may  have  copies  of 
Uie  older  editions  of  the  "Applied  Mei-hanics." 


713771 


INTEGKATIOK 


The  following  integrals  are  of  frequent  occurrence,  and  are 
here  given  for  future  reference  : 

(A)  /  d  x    V«2  —  ** 

In  the  general  formula  for  integrating  by  parts,      /  u  d  v  = 

I  v  .  d  u,  let  u  =     V«'2  —  %\   and  d  x  =   d  v  ;  then 

x  dx 

d  u  =  —       —  and  x  =  v 

t/a*  —  x' 


.-.  J  d  x    V«*  —  x-    =     x    A/"'   -  x2    +  /  - 

v     <y   a  —  x 

Again, 

/'  , f»d  x  (a2  —  a;2)  /»    x2  d  x 

.'.     fax    Va    —  ar  ==     /  — ; —  —  =   —      /  — , 

J  ^/  d1  —  x*  J  */a*  —  x3 

/' 


4. 


a?  d  x 


,-•     a-  a  x  ft          \  a  /•  ~. 

/  -  -  =  /  a2—       ==  =  a2  sin-1  _fL 

t/    yV  —  x2       t/       A/1  —  aj  a 


C. 


fd  x  A/a2  —  *2  =  x  V«2  -  a:*  +  «2  sin-1  —  +  C 
t/  a 


Let   Va2  +  a;2  =»—,«:  Then  a2  -j-  x*  =  22  —  2^,*  4-  rr2  or 


INTEGRATION. 


=  z-  —  2  z  x.     The  differential  of  this  equation  is  0  =    2  z 

(z-x)dz 


z  —  2  z  d  x  —  2  x  d  z  .-.   dx  = 

•z 


r    dx       (z  -  *) d z _   rdz_]nfr        r 

•    J     VV"+^  -    (z   -x)z    ~    J    —z~ 


z+Va'  +  xO  +  C  

(C)  fd  x   vHr+^i 

In  the  formula  for  integration  by  parts,      I  u  d  v  —  u  v  — 
I  v  d  u,  let    ^dl  -J-  x2   =  u  and  d  x  =  d  v.      Then  x  ^  v  and 
g^_g_  _  ^^ 

//»    x*  d  x 
d  x   Va*  +  x*  =  x  Va2  +  tf  -    /  y^      ^ .      Again, 

I   d  x     V~~ 


V  +  a-* 

A4J 

J   Va*- 


.:  2     I'd  x    Va*  +  x>  =  x  V«^  _|_  ^  +   a?     f          * 
J  J    V  a2  -j-  a;2 

/>       (?  x 
By  integral  B,  we  have  a2     /  —   =  a2  loge 

(a   +    Va*  +  x2)  +  C. 

...  fd  x  v-tf+i?=  ^L v^ip^+^, ioge (x+  V 

(D)  /*«2  d  cc    Va2  —  x2 

In  the  formula     /  udv  =  uv  —    /  v  du,  let  x  dx  Va'2  — 
=  d  v,  and  x  =  u  :  then  v  =  —  -  (a2  —  x2)     and  d  u  =  d  x 
-  («2  -  a:2)^   +  1          a;  (as  -  ^2)^ 


INTEGRATION. 


x>  d  x 


Transposing   the   last   term   of   the   last  member,   we   have 

-    fx*  d  x    Va*  —  x1  =  —  —  (a2  —  z2)^  4-  -   fa2 
?,J  3  3  </ 

From  integral  A  we  have 


This  integral,  taken  between  the  limits  a  and  0,  reduces  to 

— -     sin"   — ,    which  can  easily  be  memorized. 
o  (i 

(E)  /"cos2  0  d  6. 

Integrals  of  the  powers  of  sin  0  and  cos  0  are  found  by  sub- 
stituting for  these  functions  of  0,  their'values  in  terms  of  the 
multiple  angles,  as  in  the  following  case : 

//>l  _u  cos  2  0                0         sin  2  0 
cos2  0  d  0  =      /  — • d  0  = I-  • 4-  C. 
t/             2                           24 

ARTICLE  83.* 

In  solving  the  following  problems  the  student  should  always 
sketch  a  figure  representing  the  surface  or  solid  under  consid- 
eration, and  one  of  the  elementary  parts  into  which  it  is  conceived 
to  be  divided,  and  determine  the  limits  for  the  integral  by  an 
inspection  of  this  figure.  In  double  or  triple  integrations,  we 

NOTE.— The  first  three  integrals  are  taken  from  TodhnnterV  Integral  Calculus. 
NOTE.— Tl.c;  abbreviation  A.  M.  will  be  u^ed  for  Itaiikino's  Applied  ilechauics.  • 
*  The  reference  in  to  Article  S3  of  tlm  A^ied  Mechanics. 


ARTICLE    83. 


must,  in  general,  first  integrate  with  respect  to  one  of  the  vari- 
ables, between  the  proper,  limits,  and  express  the  result  in  terms  of 
the  other  variables.  This  may  be  continued  until  the  complete 
integral  is  obtained. 

To  illustrate  this  process  take  the  general  formulas  for  center 
of  gravity  of  a  solid,  viz: 


I     I     I   x  dx'dy  dz 
fff  Jx  <ly  ft 


and  similar  values  for  ?/0  and  zfl. 


The  following  application  of  these  formulas  is  found  in  Tod- 
hunter's  Analytical  Statics. 

Problem.  Find  the  center  of  gravity  of  the  eighth  part  of  an 
ellipsoid  cut  off  by  the  three  principal  planes. 

Let  Fig.  (1)  represent  the  solid  in  question,  the  equation  of 


the  surface  being  — 1-  -y,- 


If  we  integrate 
first  with  respect 
to  z,  between  the 
limits  z,  and  zero, 
we  include  all  the 
elements  (dx  dy  dz) 
in  the  prism  P  Q. 
Next  integrate 
with  respect  to  y 
between  the  limits 
y,  and  zero.  AVe 
thus  include  all 
the  prisms  in  the 
slice  between  the 
planes  ?  L  and 

mM. 

F  ro m  Equa- 
tion (1),  £5  =  C 


0) 


/           x2          if 
1 2 -~-  ;  from  Equati 


ion  of  ellipse  in  the  plane  X  Y, 


ARTICLE    83. 


/        /    '    /    '  x  dx  dy  dz         I        j        x  ax  d'y  zf 
fo  J^"1  y^     dx  dy  dz    "  y"  y*1    dx  dy  Zf 

c  x  dx  dy  1/TT^IT'        >     A  x  ,&  dy 
o  a2        b*        t/    o  «/    o    b 


/'»a         /»#,  / —        /*a        /<//!     1 

flcfyjy   1/1_^_  ^  /     /   ' 

.    J    o     .  aj  j,  J    oJ    o     I 

7"         /"*  x  ,fe  ^  /'"  «  ,fe  -,  («"  ~  O 

i/o  *         e/o  ^ 

/*"     7         ""      2//  /'"     7         ^2    /     2  2 

,/   a'"*          4  t/    .^O'" 

The  other  coordinates  of  the  center  of  gravity  will  easily  be 
found  by  taking  moments  with  reference  to  the  axes  OX  and 
OZ,  and  following  the  integration  as  above  indicated. 

PROBLEM  III.  Draw  a  line  through  C  parallel  to  E  B,  form- 
ing a  parallelogram  and  a  triangle.  A  line  joining  C  and  the 
center  of  gravity  of  the  triangle  will  be  parallel  to  O  D. 

Let  /  BOD  =  6.  Then  equating  the  moment  of  the  trapezoid, 
about  A  B,  to  the  moment  of  this  parallelogram  and  triangle, 
about  the  same  line,  we  have 

— — — -.    O  D.  sin  0  .  x0  sin  0  .  w  =  b  OD  sin  0  . sin  0  .  iv  -)- 

~OD 


3  (B+i) 

/3  (B+5)  -  (B-&)\       TKD  /     _    E-b  \ 
\         3  (B  +  i)          /  2     \          3  (B+i)/' 

PROBLEM  V.  Taking  moments  about  0  Y  we  have  for  the 
nrm  of  the  element  ydx,  the  distance  x  .  sin  Y  0  X.  The  dis- 
tance of  the  center  of  gravity  from  O  Y  on  a  line  at  right  angles 
to  0  Y  is  x0  .  sin  Y  O  X 

itx\  f'x\ 

I      xydx  I      xydx 

.  -.  x0  sin  Y  O  X  =  sin  Y  0  X  J--^ 


/"*!  f*x\ 

I      y  dx  y  dx 

t/     o  t/     o 


ARTICLE    83. 


f*x.  /«:, 

y  dx         2  /       ydx 

I/O  I/O 

The  equation  of  the  curve  is  y2  =  2  p1  x. 

PROBLEM  VII.  In  this,  as  in  many  problems  relating  to  the 
circle,  it  is  convenient  to  use  polar  coordinates.  Let  0  (Fig.  32 
A.  M.)  be  the  pole,  p  the  radius  vector,  and  0  the  variable  angle 
measured  from  the  axis  0  X.  The  circle  is  divided  into  differ- 
ential areas,  by  concentric  circles  dp  apart,  and  by  radii  making 
an  angle  cfy>  with  each  other. 

Then  a  differential  area  =  p  d  0  .  d  p. 

Its  distance  from  0  Y    =  p  cos  </>. 

Its  moment  about  0  Y   —  w  p2  d  p  .  cos  $  d  0. 
.-.  in  this  case 


/>B         />+<*> 

I        I        p  ((  p  • 
tJ    o  »'    —  4> 


+* 


^  =  _2_  R  /si 
~  3       " 


sin  0-  sin  (-0) 


(0) 

/""  xy  dx 

PROBLEM  VIIT.     x0= 


/  y  d  cc         /""         (y-2  —  ic2)a  rf  a; 

t/     r  cos  0  i/      r  cosB 

r  r     ydydx    r 

J    J     rcosfl  ___  _  t/    r 

r  r      dy  dx 

«/    O*/      r  cos  9 


cos  6  2 


PROBLEM  XVI.  Take  moments  about  0  Y.  The  moment  of 
the  annular  wedge  is  equal  to  the  moment  of  the  whole  wedge, 
minus  the  moment  of  the  interior  wedge.  (See  Art.  76  A.  M.) 

Compare  with  Problem  15. 

PROBLEMS  XIX  and  XX.  Conceive  a  plane  passing  through 
O  X,  (Fig.  39,  A.  M.)  making  a  variable  angle  y  with  the  plane 
X  Y.  Let  p  be  a  radius  vector  in  this  plane,  making  a  variable 
angle  0  with  O  X. 

Then  pd<t>dp  is  an  elementary  area  in  the  above  described 
plane.  When  that  plane  revolves  about  O  X,  through  the  angle 


10  ARTICLE    83. 

dy,  this  elementary  area  describes  a  volume  which  is  measured 
by  the  area  times  the  distance  through  which  it  moves.  This 
distance  is  the  arc  of  a  circle  whose  radius  is  p  sin  0,  and  angle 
dy,  and  =  p  sin  <pdy. 

.  •.  Elementary  volume  =  pcfy>  .  dp  .  p  sin  fydy. 

Its  distance  from  the  plane  Y  Z  =  p  cos  0;  0  Z  being  the 
third  rectangular  axis. 

/    /    /  p  d  0  .  d p  .  p  sin  cf>  dy  .  p  cos  0  .  w 

" 

(j)  .  d  p  .  p  sin  0  d  y  .  w 

j    I     I  p*  d  p.  sin  $  cos  <j>  d  <f>.  d  y 


j    I    I  p2  d  p.  sin  $  d  0.  d  y 

Integrating  between  proper  limits,  this  formula  will  give  the 
value  of  x0  for  these,  and  similar  problems. 

)    r  ~\    /J  )    2n- 

In  problem  XIX  the  limits  are  p  5-    ;      0  y    •     y  y 

XX       «  «  pj/  0|f; 

In  problem  XX,  and  in  similar  problems, 

/    /    I  pd<f> .  d  p  .  p  sin  (j)dy  .  p  sin  ^  cos  y  .  w 

MO  —     7»   ~7*   7» 

/    /    j  p  d  (f> .  d  p  .  p  sin  <p  d  y  .  w 

The  limits  for  this  integral,  in  problem  XX,  are  as  stated 
above. 

ARTICLE  91. 

The  expressions  for  moments  around  0  X,  0  Y,  and  O  Z,  on 
page  74,  are  easily  obtained  by  resolving  the  force  P,  having 
its  point  of  application  in  the  plane  X  Y,  into  components  par- 
allel to  the  three  axes,  and  taking  the  moments  of  these  com- 
ponents. Equation  4  is  reduced  by  the  relation  cos2  a  -f-  cos2  ft 
-j-  cos2  y  =  1.  See  Church's  Analytical  Geometry,  Art.  48, 
Equa.  4. 

Equation  5  A  has  for  its  first  member  the  sum  of  the  rectan- 
gles of  the  cosines  of  the  angles  which  two  lines  make  with 
three  rectangular  axes,  which  is  equal  to  the  cosine  of  the  angle 
between  the  two  lines.  See  Church's  Analytical  Geometry, 
Art.  48,  Equa.  5.  If  this  cosine  is  0,  the  angle  must  be  90°. 
The  equation  is  verified  by  multiplying  the  first  of  equations  5 


ARTICLE    95.  11 

by  cos  a,  the  second  by  cos  /3,  and  the  third  by  cos  y,  and  add- 
ing the  products.  The  second  member  of  this  sum  must  then 
be  reduced  by  introducing  the  values  of  M,,  M^  M3,  from  Equa- 
tion 3. 

ARTICLE  95. 

Equations  3.  —  The  point  whose  coordinates  are  expressed  by 
this  equation,  is  situated  in  the  angle  Y1  0  X1. 

PROBLEM. 

Find  the  moment  of  inertia  of  a  rectangle  of  length  h  and 
breadth  5,  about  a  neutral  axis  perpendicular  to  its  plane.  See 
Equation  5. 


Equations  13.  —  Complete  figure  45,  A.  M.,  by  drawing  a  line 
from  C,  parallel  to  0  T,  and  call  the  point  where  this  line  inter- 
sects the  line  O  Xn  N.  From  N,  draw  a  line  parallel  to  T  C, 
and  call  its  point  of  intersection  with  0  T,  H.  The  angle 
T  0  Xj  =  90°  —  /  X,  0  y  =  /31. 

Take  the  most  general  form  of  the  equation  of'  a  straight 

line,  A»-|-B?/-|-C=iO,  and  write  it  in  the  form  -~-  x  +7T 

y  -\-  1  =  0.     Since  this  equation  is  true  for  all  values  of  x  and  y, 
it  will  be  true  when  x  or  y  becomes  0. 

Let  the  intercept  on  the  axis  0  X  =  c,  and  the  intercept  on 
0  Y  =  d,  so  that,  when  y  =  0,  x  =  c,  and  when  x  =.  0,  y  =  d. 

Make  y  =  0.     Then  -^  c  +  1  =  0,  or  -^-  =  - 

B  B  1 

Make  x  —  0.     Then  -^-d+l  =  0,  or  -^  =  —  ~^ 

These  values  introduced  into  the  general  equation,  reduce  it 


.  (A) 

Now  multiply  both  members  of  this  equation  by  n,  drawn 
perpendicular  to  the  straight  line,  and  making  the  angle  /31  with 
the  axis  O  X.  See  line  O  T,  fig.  45,  A.  M. 


12  ARTICLE    95. 

Then    —  x  -j-  —r  y  =  n.     But  —  =  cos  /31  and  -y-  =  sin/31 

.-.  x  cos  /31  +  y  sin  /31  =  w.  (B) 

The  equation  of  a  tangent  to  an  ellipse  (T  C)  is  a2  y  y1  -f  J*  a; 

x1  =  a2  62  (see  Olney's  General  Geometry,  Art.  136) 

—  .  --  1_  jL  —  i.     Multiply  this  equation  by  n. 


By  comparing  equations  C  and  B,  it  will  be  seen  that  cos  /31, 

x1  n  y1  n 

-^5-,  and  am  0'  =    *_ 

a2  cos  /31  &2  .  sin  /31 

.-.  ^  =  —  —  ^-,  and  f  =  -  ^-  -  (D) 

Substitute  these  values  in  equation  B,  for  x  and  y. 


n 

To  find  the  value  of  n  t. 
The  general  equation  of  a  normal  to  an  ellipse  is  y  —  y1  = 

a*  y1 

JJTJA(X  —  a1).     (See  Olney's  General  Geometry,  Art.  156). 

In  this  equation  make  y  =  0,  and  divide  both  members  by  y1. 
-   1  =   JL_   (x   _   ai);    ...  _  y  x1  =  a2  x  —  a2  a1;  x  — 

fr*'  ~o,^  gl  =  O  N. 

From  Equation  D,  we  have  a;1  ==  —     — —  • 

...  0  N  =    (gl-y)<™/31 

w 


.-.%<=  (a2  —  &2)  cos  /31 .  sin  /31. 

PROBLEM  IY. 

Use  polar  coordinates.      An  elementary  area  =  p  d  <f>  .  d  p  . 
(See  Note  on  Problem  VII  of  Art.   83.)     Its  distance  from 


ARTICLES    102-112.  13 

0  Y  =  p  cos  0.     Its  moment  of   inertia  =  p  d  ^  d  p  (p  cos  </>)2 


gral  E). 

ARTICLE  102. 

Equations  1  and  2  are  given  in  Ch.au  venet's  Trigonometry, 
page  163,  formula  53,  and  page  162,  formula  48. 

ARTICLE  105. 

PROBLEM  II.  —  According  to  the  preceding  notation,  the  form- 
ulae near  the  top  of  page  91  A.  M.  should  be 
O  D  =  pxx  .  area  0  B  C  -f  pyt  .  area  0  C  A  -f  pzx  .  area  0  A  B. 
OE=pff.  area  0  B  C  -j-  p     .  area  OCA  +  p,v.  area  0  A  B. 
O  F  =  pxz  .  area  0  B  C  +  pyz  .  area  0  C  A  -f  pzz  .  area  0  A  B. 

ARTICLES  110  AND  111. 

The  student  should  carefully  study  these  articles,  and  mem- 
orize the  theorems,  in  order  to  understand  Art.  112. 

ARTICLE  112. 

Equation  1.  —  From  figure  54,  A.  M.,  we  have 
0  R2  =  OM*  +  MR*  -f  2  CTST.  M^TTcos  R  M  N 


Px  +  A2  Px  —  P?  A  „  2  A 

=          2  +  2  cos  2  a;  w   =  ^   (1  +  cos  2  x  n) 

+  ^  (1  -  cos  2  x  n). 

A  A  A 

But  cos  2  a;  n  =  2  cos2  icw—  1  =  1  —  2  sin2  x  n. 

A  A 

.-.  Pr  =  P?  cos2  x  n  4-  ^y2  sin2  a;  » 

4/J  A  A    ) 

•'•  Pr  =    '   \  P*  •  cos2  x  n  +  pf  sin2  x  n  [  =  0  R.  (1) 

Equation  2.  —  From  the  triangle  M  0  R,  we  have 
sin  N  O  R  :  sin  R  M  0  :  :  M~R  :  O~R 


14  ARTICLE    112. 


or  sin  N  0  R  :  sin  (180  —  2  x  n)  ;  ;  Px  ~  Py  :  Pr 

A  A    p  _  p 

.:  sin  N  0  R  =  sin  n  r  =  sin  2  x  n  —  -  — 


or  N  0  R  =  n  r  =  arc  .  sin  (sin  2  x  n  Pl~  Pv\  •  (2) 

Proof  that  the  locus  of  the  point  K  is  an  ellipse  whose  semi- 
axes  are  px  and  pf. 

From  the  construction  of  figure  54,  of  the  Applied  Mechanics, 
it  is  readily  seen  that  since  OM  =  MQ  =  MP,  the  line 

PR  =  PM  -  MR  =  Pi+Pi-P^  =py. 

Also  QR  =  QM  +  MR  =  p*  +  p*  +  ^^  =  Px. 

A 
The  angle  OQM  =  90°  -  xn, 

Therefore  for  the  coordinates  of  the  point  R  we  have 

A  A 

x  —  QR  cos  xn  =px  cos  xn. 

A  A 

y  =  PR  sin  xn  =  py  sin  xn, 

Multiplying  the  first  of  these  equations  by  py,  the  second  by  px; 
squaring  and  adding  we  have 

Plx*+Plf=plpl 
which  is  the  equation  of  an  ellipse  with  semiaxes  px  and  pv. 


ARTICLE    112.  15 

Equations  3  and  4. 
pn  =  OM  -  MR  cos  (180°  -  2  CT)  =  &L±P»+P*~P* cos  2  zn 

—  ^  (1  -f  cos  2  on)  +  ^  (1  —  cos  2  xn)  =px  cos2  z»  + 

A 
py  sin2  £«. 

j»t  =  MR  sin  2  o:»  =  ^  ~  Py.  2  sin  xn  cos  xn  =  (px  —  py)  sin  xn 

A 
cos  xn. 

Comparison  of  figures  54  and  57,  A.  M. 

In  figure  54  suppose  a  second  plane  making  an  angle  -with 
the  plane  A  B,  to  be  drawn  through  0. 

Let  the  same  construction  be  made  to  represent  the  compo- 
nents of  the  principal  stresses  on  this  plane,  as  is  made  for  AB. 

The  distance  set  off  on  the  normal  to  this  supposed  plane, 
must  be  equal  to  OM,  and  the  line  corresponding  to  M'R,  must 
be  equal  to  MR.  There  will  also  be  a  line  corresponding  to  0  R. 
Now  suppose  this  second  system  of  lines  to  be  revolved  about  0, 
until  the  plane  coincides  with  AB.  The  result  will  be  ORMR'O 
of  figure  57.  By  the  construction  of  figure  57  it  will  be  seen 
that  MR  =  MR'  and  OM  is  the  same  for  both  systems  of  lines, 
hence  the  figure  satisfies  the  condition  of  the  problem-,  viz  :  that 

OM  =?£±^and  MR  =  ^-L» . 
2  2 

Equation  15. — From  the  triangles  OMR  and  OMRr  figure  57. 
MR2  =  (JM2  +  jt;2  —  2  ^OM  cos  nr. 
MR'2  =  OM2  +  p1'1  -  2  p'  OM  cos  $/. 
.-.  0  =  p2  —  p'2  —  2.  OM  (p.  cos  nr  —  p1  cos  rcV). 

X  -  X2 

•'•  OM  ==  -       — —  (15) 

2  (p.  cos  nr  —p'  cos  n V) 


16  ARTICLE    112. 

Equations  16,  are  found  by  substituting  t^LJLjOL  for  OM  in 
the  above  values  for  MR,  and  MR'. 

/I/nations  17.  —  For  the  first  of  these  equations,  draw  a  line 
from  R  perpendicular  to  ON  (Fig.  57)  and  call  its  intersection 
with  ON,  A. 

Then  cos  2  xn  —  cos  NMR  =  —^  .  MR  =  ^L-Jjf  . 

MR  2 

MA  =  0  A  —  OM  =  OR  cos  NOR  —  OM  =  p  cos  nr  —  J~^'- 

A 
_  2  p  cos  nr  —  px  —  py 

2 

A 

n  A          2  »  .  cos  nr  —  p    —  p 
.:  cos  2  aj»  —  —  -  £=  —  £*  . 

PX~Py 

A 

The  value  of  cos  2  xr/  is  found  by  drawing  a  perpendicular  to 
ON,  from  R',  and  repeating  the  above  process. 

Equation  27. 

From  equation  25,  p  +  p'  =  V/V^°^  .  (A) 

COS  (ft 

Write  equation  25  in  the  form 

A 

cos2  0  (jo2  -f-  2  _p_?/  -j-  ;/2)  =  4  7>  ;/  cos2  «r,  suotract  4  pp*  cos2  ^> 
from  both  members  of  the  equation,  and  extract  the  square 
root. 


p  —p'  =  2  V/'  /•  V  cos2  wr  —   cos2  0  (B) 


Then .       .          __^_ 

COS  (p 

From  equations  A  and  B,  we  find 

/ A  A 

|/ p  p'  (  cos  nr  -f-  V(  cos2  «r  —  cos2 


A  A 

,  _  '  (cosnr  —  v(cos2  nr  —  cos2 

cos  0 

A  A 


„/       cos  7/r  —  v(cos2  nr  —  cos2  ^>) 


cos  «?•  -j-  V(  cos2  nr-—  cos2 


ARTICLE     124.  17 

ARTICLE  124. 
Formula  for  the  center  of  pressure  of  a  fluid,  upon  a  plane  surface, 

In  figure  62,  let  BF  be  the  plane  and  suppose  FO  to  be  a  line 
representing  the  continuation  of  this  plane  to  O,  the  surface  of 
the  fluid.  Let  OB  be  the  axis  of  x,  and  the  axis  of  y  at  right 
angles  to  it. 

Let  the  angle  that  OB  makes  with  the  surface  of  the  fluid  be 
represented  by  Q. 

Let  F  be  the  origin  of  coordinates,  and  x  be  measured  pos- 
itively downward. 

dxdy  =  differential  portion  of  this  plane. 

The  pressure  on  a  unit's  surface  of  this  plane  is  normal  to 
the  surface,  and  measured  by  p  =  w  (x  -|-  OF)  sin  0,  in  which 
w  is  the  weight  of  a  unit  volume  of  the  fluid.  (See  Art.  110, 
A.  M.) 

Let  OF  =  a.     Then  p  =  w  (a  -f  x)  sin  Q. 

By  formula  2  of  Art.  89,  A.  M., 

/>-fy  /JFB 

/  /       w  x  (a  -\-  x)  sin  6  .  dxdy 

x°  =  -/»+?   '  /•*'" 

I  I         w  (a  -}-  a:)  sin  Q  dxdy 

/4y         /»FB 
_y   J  o(ax  +  x>}dxdy 

~  jZf"(a+^' 
(a  +  x) 


/•     y  /»FB 

/  /         dxdy  (a  +  x) 

t/      -y    t/      o 

e  surfac 
uations 

/-»FB 

/         y  (ax  -(-  a;2)  dx 


If  the  plane  surface  is  symmetrical  with  respect  to  the  axis  x, 
the  above  equations  become 


f" 


(B) 

If  the  plane  extends  to  the  surface  of  the  fluid,  a  becomes  zero, 

/     x2  ydx 
and  the  equations  B  reduce  to  x0  =  - — ^ —    — ,  h  being  the 

depth  of  the  plane  surface.  (C) 


18  ARTICLES    124-159. 

PROBLEM  1. — Find  the  position  of  the  center  of  pressure  on 
a  rectangular  plane  surface  whose  depth  is  h,  and  breadth  I, 
when  its  upper  edge  is  parallel  to  the  surface  of  the  fluid,  and 
at  a  distance  a  below  it. 

From  equation  B  we  have       * 

y»A 
y  (ax  -}-  x2)d  x 

x  =  _ \—  But  y  =  -  b 

y  (a  +  x)  d  x 


6  ah 


PROBLEM  2. — Find  the  centers  of  pressure  of  the  following 
surfaces,  when  their  bases  are  in  the  surface  of  the  fluid.  (The 
depth  of  each  surface  =  h.)  Rectangle,  triangle,  and  parabola. 

214 

Ans.  xn  =  -  h,  -  h,  -  h,  respectively. 
o        2t        i 

PROBLEM  3. — The  parabola  with  its  vertex  in  the  surface,  and 
base  parallel  to  it. 

Ans.  XH  =  -^  h. 

PROBLEM  4. — Find  the  center  of  pressure  on  a  triangular  sur- 
face whose  depth  is  A,  base  parallel  to  the  surface  of  the  fluid, 
vertex  upward  and  at  a  distance  a  below  the  surface  of  the  fluid. 

4  a  It*  4-  3  /i3 

Ans.  xu  =  -z  T  —  distance  from  vertex. 

6  a  h   -}-  4  h2 

ARTICLE  159. 

w 

Equations  4. — The  load  —  at  5,  is  resolved  into  two  compo- 
nents, —  cos  i,  which  is  transmitted  through  5  4  and  produces 
a  stress  on  all  the  pieces  of  the  secondary  truss  1534,  and 
—  sin  i,  one-half  of  which  is  supposed  to  cause  a  pull  on  the 

piece  1  5,  and  one-half  a  thrust  on  the  piece  5  3. 

Equations  4  are  obtained  by  computing  the  stresses  due  to  a 


*  NOTE.— It  will  be  noticed  that  ar0  is  measured  downward  from  the  top  of  the  plane, 
and  not  from  the  surface  of  t/te  fluid.    Also  that  a  is  measured  on  the  axis  of  x. 


ARTICLE    159.  19 

w 
force    —   cos  i  acting  in  the  direction  54,  and  combining  these 

w 
stresses  with  the  component  —  sin  i  above  mentioned. 

Draw  a  line  parallel  to  54  and  equal  to  —  cos  i.      From  its 

extremities,  draw  lines  parallel  to  14  and  34,  and  from  their 
intersection  draw  a  line  parallel  to  13,  to  meet  the  first  line,  to 
which  it  is  perpendicular. 

W  W 

Then  from  this  diagram  R4,  =  R41  =  —  cos  i,  cosec  i  —  — 

o  o 

cotan  i. 

R,5  =  —  cos  i  cotan  i  (from  diagram)  -\-   -  w  sin  i  (see  above) 

w    /cos'2  i  -4-  sin'2  «'\          w 
s±  — -  -I  —  — r-1--       - 1   —   —  cosec  t. 
8    \         sin  i         /          8 

R51  =  —  cos  i  cotan  i  (from  diagram)  —  ^  w  sin  i  (see  above) 

w     /I  —  sin-  i  —  sin'2  i\  w     / 1  —  2  sin2  i\  w 

IT    V  sin  i  /    =  :    ~8~    V        sml       /    =       T 

(cosec  i  —  2  sin  i). 

The  Method  of  Drawing  Reciprocal  Diagrams. 

Since  the  forces  acting  through  any  joint  of  a  truss  must  be 
in  equilibrium,  these  forces  may  be  represented  by  a  closed 
polygon  whose  sides  are  parallel  to  their  directions. 

A  complete  diagram  for  a  truss  under  a  given  load  ought, 
therefore,  to  contain  as  many  such  polygons  as  there  are  joints 
in  the  structure.  In  most  cases  a  variety  of  diagrams  can  be 
drawn,  any  one  of  which  will  correctly  represent  the  stresses  in 
a  given  truss.  The  one  that  fulfills  most  nearly  the  condition, 
that  for  each  joint  in  the  truss,  there  shall  be  in  the  diagram 
a  closed  polygon  whose  sides  are  parallel  to  the  forces  acting 
through  that  joint,  and  taken  in  the  same  order  as  those  forces, 
has  been  called  the  reciprocal  diagram. 

A  method  of  notation  given  by  R  H.  Bow,  C.  E  ,  renders  the 
construction  of  reciprocal  diagrams  very  easy  in  most  cases. 
This  method  consists  in  placing  upon  the  figure  of  the  truss,  a 
letter  in  each  of  the  angular  spaces  formed  by  the  intersection 
of  the  forces  at  each  joint  (a  single  letter  in  an  enclosed  area 
answering  for  all  the  internal  angles  formed  by  its  sides),  and 


20 


RECIPROCAL    DIAGRAMS. 


in  the  diagram  the  same  letter  at  the  junction  of  two  lines  rep- 
resenting concurring  forces,  that  is  found  in  the  angle  between 
those  forces  upon  the  figure  of  the  truss. 

To  illustrate  the  application  of  this  notation,  let  us  construct 
the  diagram  for  the  frame  represented  by  Fig.  78  A.  M.,  on 
the  supposition  that  it  is  loaded  with  W  uniformly  distributed 
over  12  and  13. 

1st.  Distribute  the  load  upon  the  joints,  as  directed  in  Art. 
147  A.  M.,  or  for  a  uniformly  distributed  load  by  placing  upon 
each  joint  half  of  the  load  between  the  adjacent  joints.  This 

distribution  gives  -  W  at  the  joints  4.  6.  1,  8  and  10,  and  — W 

at  the  joints  2  and  3. 

2d.  Represent  the  directions  of 
the  external  forces  by  lines,  as  in 
the  accompanying  Fig.  1.  The 
forces  are  all  given  in  direction, 
the  internal  forces  or  stresses  be- 
ing represented  in  direction  by 
the  pieces  of  the  frame. 

3d.  Letter  the  angular  spaces 
formed  by  concurring  forces. 

In  Fig.  1,  P  occupies 
K  the  space  between  3*5 
a  and  the  left  hand  sup- 
port, 3  5  and  5  7,  5  7  and 
7  9,  7  9  and  9  2,  and  9  2 
and  the  right  hand  sup- 
port. K  occupies  the 
angular  space  at  the  left 
of  the  left  hand  support, 
between  it  and  the  load 
at  3.  A  occupies  the 
angular  space  between 
3  5  and  3  4,  3  4  and  4  5, 
and  4  5  and  5  3,  and  so  on. 
4th.  Draw  a  line  of 
loads  (vertical  in  this 
case)  representing  the 
whole  load  W,  and  di- 
vide it  into  parts  repre- 
senting the  external 
forces  acting  at  t  h  e 
Fig.  2.  joints. 


Fig.  1. 


RECIPROCAL    DIAGRAMS.  21 

As  the  supporting  forces  here  are  vertical  and  equal,  they 
will  each  be  represented  by  half  this  line  of  loads  (K  P  and 
L  P,  Fig.  2). 

5th.  Letter  the  line  of  loads  to  correspond  with  the  lettering 
of  the  truss.  In  reading  from  K  to  «,  Fig.  1,  we  cross  the  line 

W 

representing  the  load  —  at  3.     Therefore  K  a  on  the  line  of 

"W  W 

loads,  must  be  —  •      For  a  similar  reason  a  b  =  -  -,  and  so  on. 

Since  the  loads  at  2  and  3  rest  directly  and  vertically  upon 
the  supports  of  the  truss  (and  the  supporting  forces  are  vertical), 
they  produce  no  stress  on  the  frame,  and  may  be  left  out  of 
account. 

Then  a  h  upon  the  line  of  loads  will  represent  the  effective 
load,  and  a  P  and  h  P  the  supporting  forces  to  sustain  that  load. 
The  loads  at  2  and  3,  and  the  letters  K  and  L  could,  therefore, 
be  omitted  in  Figs.  1  and  2. 

6th.  Commence  at  some  joint  where  all  but  two  of  the  forces 
are  known,  draw  the  polygon  of  forces  for  that  joint,  and  by 
proceeding  from  one  joint  to  another  complete  the  diagram. 

Commencing  at  the  left  hand  joint  of  the  truss  Fig.  1,  and 

"W 

leaving  out  the  load  —  as  above  directed,  v/e  have  on  the  line 

1  i 

of  loads,  Fig.  2,  the  distance  a  P,  upon  which  to  construct  a  tri- 
angle having  its  sides  parallel  to  3  4  and  3  5. 

Draw  through  «,  Fig.  2,  a  line  parallel  to  3  4,  Fig.  1,  and 
through  P  a  line  parallel  to  3  5,  and  mark  their  intersection  by 
the  letter  A. 

Then  pass  to  the  joint  at  4,  draw  lines  through  A  and  b  of 
Fig.  2,  parallel  to  4  5  and  4  G  of  Fig.  1,  and  mark  their  inter- 
section with  B.  Through  B  draw  a  line  parallel  to  5  G  to  meet 
a  line  through  P,  parallel  to  5  7,  and  mark  the  intersection  C.  In 
a  similar  manner  complete  the  diagram  represented  by  Fig.  2. 

The  advantage  o^f  this  notation  lies  in  the  fact  that,  if  the  let- 
ters in  the  angular  spaces  about  any  joint  are  taken  in  order, 
and  the  same  letters,  taken  in  the  same  order,  are  found  in  the 
diagram,  at  the  vertices  of  the  polygon  representing  the  forces 
at  that  joint,  the  diagram  is  correct.  Thus  in  Fig.  1,  for  the 
joint  3,  we  have  aAPa,  and  in  Fig.  2,  aAPa  forms  a  closed 
polygon.  Reading  about  joint  1  of  the  frame  we  have  dDJ&cd, 
and  t/DEeo7  forms  a  polygon  in  the  diagram. 

Failing   Cases. — Whenever  we  find  on  arriving  at  any  joint, 


22 


RECIPROCAL    DIAGRAMS. 


more  than  two  unknown  forces,  this  method  apparently  fails, 
since  an  indefinite  number  of  polygons  may  then  be  drawn  with 
corresponding  parallel  sides.  A  careful  inspection  of  the  truss, 
in  connection  with  a  sketch  of  the  general  outline  of  the  diagram, 
will  often  reveal  some  condition  fixing  the  relations  of  the  lines 
in  the  diagram. 

This  is  the  case  in  Fig.  77,  A.  M.  Arriving  at  joint  5,  we  find 
the  forces  parallel  to  5  4,  5  7,  and  5  1  unknown  ;  but  by  sketching 
a  diagram  for  this  joint,  we  see  that  the  lines  representing  the 
stresses  on  4  7  and  1  7  must  overlap  or  be  portions  of  the  same 


Fig.  3. 


Diagram  of  stresses  for  half  the  trass  represented  lij  Fig.  77,  A.  M. 
Letter  that  figure  in  a  manner  illustrated  by  Fig.  1. 

line,  in  order  to  read  correctly  for  the  joints  4  and  7  ;  and  since 
5  7  and  7  1  make  equal  angles  with  1  3,  the  lines  on  the  diagram 
representing  their  stresses  must  intersect  midway  between  lines 
drawn  through  e  and/,  parallel  to  1  3.  Hence,  as  CD  is  the  line 
to  be  drawn,  we  may  draw  the  line  from  C,  parallel  to  5  4,  to  a 
point  D,  half  way  between  the  parallel  lines  above  mentioned. 

There  will  be  no  difficulty  with  the  remaining  part  of  the 
problem. 

Or  otherwise  :  Since  the  stresses  on  5  7  and  5  6  are  seen  to  be 
equal,  and  make  equal  angles  with  the  rafter,  we  may  draw  an 
auxiliary  line  from  B  parallel  to  5  4,  till  it  meets  a  line  through 

NOTE.— For  an  extended  discussion  and  application  of  these  principles,  the  student 
is  referred  to  "  Economics  of  Construction,''  by  II.  II.  I'.ow,  C.  E. 


ARTICLES    160-169.  23 

e  parallel  to  the  rafter,  which  gives  E.  From  E  and  C  draw 
lines  parallel  to  5  7  and  5  4,  which  will  intersect  in  D.  It  will  be 
seen  that,  this  makes  ED  =  BC,  or  the  stress  upon  5  7  equal  to  the 
stress  upon  5  6. 

PROBLEMS.  —  Draw  diagrams  for  figures  79  and  82,  A.  M. 

ARTICLE  160. 

EXAMPLE  1.     The  T67  of  the  first,  and  T45  of  the  last  equations 
at  the  end  of  this  example,  should  bo  omitted. 

ARTICLE  167. 

The  funicular  polygon  is  defined  in  Article  152  instead  of 
Article  150. 

ARTICLE  169. 

Equation  8.  —  Combine  the  equation  a^  -{-  x.2  =.  a,  with  the 
equation  obtained  from  the  proportion  yv  :  7/2  '  *  x\  ;  x\. 

x  2        x2 

Equation  9.     See  equation  5  ;  in  =  -L  =—  ;    xf  =  4  my  v 

4yi      4y2 

2      __      ,2 

x\  =  4  m  y2   .-.  x*  -f  a:'  =  4  m  (y,  +  yt),  .:  m  = 


4  (y,  4-  y2) 

The  remaining  part  of  the  equation  can  be  obtained  by  sub- 
stituting the  values  of  xl  and  x.2  in  equation  8. 

=      fds  =       C  A/  d  x*  +  d  if   =     fdy 

dy>  '  _ 

From  the  equation  of  the  curve.  «!  =  4  m  y;  —~=- — =A/^L 

dy       x        '    y  ' 

.:*=f<l!>     .          y 
Then  by  Integral  C,  we  have 

^r  loSe 


=  v/y  vwi  +  y  +  m  ioge  (  vy  +  vin  +  y  )  +  c. 

When  y  =  0,  s  =  0.   .-.  0  =z  ??i  loge  v/Hi  -(-  C  or  C—  —  m  \ogeVm, 


ARTICLES    169-172. 


Vy  (,n  +  y]  -j-  m  (log,  (^y  +  \/m  +  y  )  —  loge 
loge  (  V^  +  A/^JJ^/)  _ 


_«*y  _Jl  _)_  1   =  m  tan  «'  .  sec  i.     See  equation  6. 


.-.  ,9  =•  m  -<  tan  i .  sec  i  .  -j-  loge  (tan  *  -f-  sec  «') 

(m  \^ 

—  -+-  1 1   .by  the  binomial 

formula,  multiply  by  dy  and  integrate  two  terms  of  the  series. 

ARTICLE  171. 

THE  LAST  OF  EQUATIONS  1.     Place  the  letter  G  at  the  intersec- 
tion of  AB  and  EF  prolonged. 

W 

Then  DE  =  |  EF  =  £  EG !L_Y since  EF  :  EG  \\  ef 


(since  E  G  =  |  B  C,  for  A  E  =  \  A  C). 

At  the  top  of  page  171. 

Place  the  letter  A  in  its  proper  position  in  Fig.  86.  y,  is  the 
projection  of  the  line  DE  upon  a  line  drawn  perpendicular  to 
AC. 

ARTICLE  172, 

Equation  11.  In  Fig.  87,  A.  M.,  draw  lines  through  B  and 
x  perpendicular  to  the  arrow  line  through  P.  It  will  be  seen 

that  t  .  cos  i  =  -  .  cos  /  .-.  i  =  arc  .  cos  I  —  cosy)  • 

\2 1          / 


ARTICLES    172-174.  25 

Equation  12.  Equation  16  of  Art.  169,  gives  the  length  of  a 
parabola,  from  its  vertex  to  a  point  whose  tangent  makes  the 
angle  i  with  the  axis  of  x,  tangent  at  the  vertex.  Equation  1 2 
gives  the  length  of  that  portion  of  the  curve  included  between 
two  points  where  the  inclinations  are  i  and/.  This  equation  is 
obtained  directly  from  equation  16  Art.  169,  by  taking  that  in- 
tegral between  the  limits  i  and  /. 

Equation  13. — In  equation  17,  Art.  169,  substitute  x  -f  y  sin 
/  for  x,  and  y  cos/  for  y.  These  values  are  obtained  by  an  in- 
spection of  Fig.  87,  A.  M. 

AETICLE  174. 

d2  u        u       2  du    d2u       ludu         ,  /du?\ 

Equation  (a).      --- -  =  —  ;    -  — — = -—  or  d  I  —  I 

v  '        d  x2        a?  dx1  a1  \dx2/ 

du2        u?  a  d  u 


Let   v  u'2  -)-  a'  c  —  z  -[-  u 

then  u2  -f  a2  c  =  z2  -f  2  z  u  -f-  w2  or  a2  c  =  z*  +  2  z  u.         (A) 
Differentiate  and  divide  by  2. 

2  +  u 

0=  z  d  z  -\-  z  d  u  -\-  u  d  z  or  «  w  =:  —  —  •   rt  2 

z 

.-.  (?  x  =  —  a  --  —  .-.  x  =  —  a  loge  z  ~h  a  1°&:  ci-      ^n  which 
a  loge  ct  is  the  constant  of  integration. 

.•.    —    =    —   logc  Z    +    log,  Cj   =   loge    -±  • 

c,  c. 


From  equation  A  above,  we  have  u  = 


2  z 


j        „-  - 

e e       =Ae     —  Be 


2c,  2 


Equation  (c). — From  Equation  1  we  have  —. —  =  y.  Hence, 

differentiate  equation  (fc),  and  divide  by  d  x  ;  the  result  will  be 
equation  (c). 


26 


ARTICLE    175. 


Equation    6  is  derived  from  (4),   by  multiplying  by  2   ea , 
which,  reduces  it  to  the  form 


AKTICLE  175. 
Equation  4. — By  squaring  both  members  of  equation  3,  and 


solving  with  respect  to 


X  <l    X 

- , —  we  and  — -, —  = 
d  s  d  ~ 


The 


numerator  of  the  second  member  is  m,  and  not  m2. 

Equation  5. — Integrate  equation  4  by  means  of  integral  B. 
Equation  8. — Divide  both  members  of  Equation  o  by  m,  and 

X  g  / 72 

write  the  resulting  equation  in  the  form  e  m  —  —  -|-  \  1  -(-  ~i- 

Substitute  the  value  of   —  from  Equation  6,  for  the  first  term 

1  /     =L         -*\ 
of  the  last  member,  and  transpose.    The  result  is  -  I  e  »*  -(-  e    »« I 

—  |/  1  -|_  -f_  .     This  value  in  the  first  part  of  the  first  value  of 


y  (Equation  8),  gives  y  =   y^s2  +  m*   —  m. 

Equations  11.  —  Combine  xl  -J-  x2  =  h,  and  a;,  —  £2  =  ^. 

Equations  12.  —  Putting  these  values  of  .r  in  Equations  6  and 
8,  the  following  will  be  found  to  be  the  correct  results  : 

' 


y,-,^^^    ""-«    "'").    (c-    '"-,    '») 


NOTE.—  The  reference  in  the  last  line  of 
of  173.  as  in  some  editions. 


Article  174,  instead 


ARTICLE    175. 


27 


Properties  of  the  Catenary. 

I.  The  radius  of  curvature  may  be  found  by  making  proper 
substitutions  in  the  formula 


p  = 


d  y 


or  as  follows  : 


P  =   d 


in  which  ds  =  (dx*  +  di 
d  y  s 


From   Equations  1  and  3, 


=  — 
ax 


d9  y 

—  =  — 
a  x 


d  s 


dx.  ds  /  ds  V* 

d2  y  =  -  .-.  p  =  d  x  .  d  x  .  d  s  =  m  I  —  -  —  -  1     =  m  .  sec2  ^. 
m  \  d  x  ' 


II.  From  the  Equation 
y'1  =  .s2  -j-  m-, (obtained  from 
8  by  transferring  the  origin 
of  coordinates  to  a  distance 
m  below  the  vertex,)it  will 
be  seen  that,  if  a  right  tri- 
angle be  constructed,  having 
y  for  its  hypotenuse,  and  m  for  the  side  adjacent  to  the  foot  of 
the  ordinate,  the  remaining  side  will  be  s,  or  the  length  of  the 
curve  from  the  vertex  to  the  point  having  y  as  its  ordinate. 

d  y  s 

From  Equations  1  and  2.  we  have  tan  i  =  — r^-  ==  - 

a  x  m 

Hence,  if  the  lines  are  drawn  as  B  D  and  D  A  in  Fig.  4,  tan  i 

=  tan  <  A  B  D  =  =  - —  ;   hence,  the  line  A  D  is  tan- 

JJ   -B  in 

gent  to  the  curve  at  A. 

If  we  draw  A  C  perpendicular  to  A  D,  we  have  from  the 
similar  triangles  A  B  D  and  A  B  C,  in  :y\'.  y  :   A  C  .-.  A  C 

=  — -  =  m  I  -- 1  =  m  .  sec-  A  B  D  =  m  .  sec2  i  =  p.     See 
m  \  m  / 

Equation  17. 


t  NOTE. — The  line  A  D,  prolonged,  does  not  necessarily  pass  through  O. 


28 


ARTICLE    83. 


Fig.  6. 


IV.  In  any 
position  of  the 
rolling  p  a  r  - 
abola,  a  line 
drawn  from 
the  focus  to 
the  point  of 
contact,  will 
be  normal  to 
the  curve  de- 
scribed by  the 
focus,  since  it 
is  the  instan- 
taneous radius 
vector  of  the 
describing 


point.     (See  A.  M.  at  the  bottom  of  page  401.) 

The  following  shows  that  this  line  is  equal  to  m  .  sec5  i : 
Let  B,  in  Fig.  6,  be  any  point  in  a  parabola  whose  focal  dis- 
tance is  w,  F  its  focus,  and  B  E  a  tangent  at  the  point  B,  making 
the  angle  ft  with  the  axis  0  X.     The  angle  E  =  <  B  (from 
Analytical  Geometry),  =  90°  —  ft,  from  figure. 

When  the  parabola  has  rolled  till  B  comes  into  0  X  at  B1,  the 
line  B  E  will  coincide  with  ()  X,  the  line  F  B  will  be  at  F1  B1  and 
make  an  angle  =  90e  —  ft  with  0  X,  since  a  line  drawn  from  B 
to  the  focus  F  will  retain  its  position  relative  to  the  tangent, 
and  a  line  drawn  through  F1  _L  F1  B1  will  be  tangent  to  the  curve 
described  by  the  focus,  since,  as  before  stated,  F1  B1  is  normal 
to  that  curve. 
Then  ft  =  i. 
From   the  polar   equation   of   the  parabola,   we   have  p  = 

2  m 

when  0  =  <  B  F  E.      (See    Church's   Analytical 


Geometry,  pp.  128  and  123.) 
6  =  180°  —  2  (90°  —  ft)  =  2  ft 
=  cos  2  i  =  I  —  '2  sin2  i 
2m  m 


2  i 


i  =  F  B  =  F1  B1. 


~  2  (1  -  sin't)         cosH'  ~ 

Therefore  as  F1  B1  =  m  .  sec2  i,  and  is  normal  to  the  curve 
described,  it  follows  by  the  first  and  second  geometrical  prop- 
erties that  this  curve  is  a  catenary. 


ARTICLE    181.  29 

ABTICLE  181. 

The  Geometrical  Construction  preceding  Equations  5. 
At  this  point  we  need  to  show  : 
1st.     That  2  O^i  =  A  +  B. 
2d.      That  ap  and  aq  are  equal  to  A  and  B.. 
3d.      That  by  construction,  the  triangle  p  O'q  has  a  right  angle 

atO'. 

4th.     That  qp  is  perpendicular  to  O'a,  and  tangent  to  the  ellipse. 
5th.     That  the  angle  p  O'a  is  that  made  by  a  perpendicular  to 
the  tangent  qp,  with  the  major  axis  of  the  ellipse. 

1st. 

Complete  a  parallelogram  upon  O'a  and  O'B',  and  call  the  vt  r- 
tex  opposite  0',  E.  Then  by  construction,  the  point  E  is  in  tl  u 
prolongation  of  O'm,  which  is  a  diagonal  of  this  parallelogram. 

From  Trigonometry,  O'E*  =.  (2  O'm)2  =  0'1T'2  -f  0V  -f  2 
O'B' .  O'a  .  cos  B'O'a. 

By  an  examination  of  Fig.  89,  A.  M.,  it  will  be  seen  that 
/  B'O'a  =  /  j.  Since  O'A'  and  O'B'  are  conjugate  diameters 
of  the  ellipse,  and  O'a  =  O'A'  by  construction,  let  O'B'  =  A' 
and  O'a  =  B'. 

Then  we  have  from  the  preceding  equation,   (2  O'm)2  =  A'2 
-f  B'2  -f  2  A'B'  cosj. 
cosy  =  cos  .  (90°  —  A'O'B')  =  sin  of  the  angle  between  the 

conjugate  diameters  A' and  B'  =  — - /;  and  A'2  -f-  B'2  =  A2 

.A.     Jj 

-f-  B2.     See  equations  2  and  3  of  Art.  157,  Church's  Analytical 
Geometry. 


.-.  (2  O'm)'  =  A2  +  B2  +  2  AB  or  2  O'm  =  A  +  B  £ . 

2d. 

In  a  similar  manner  we  obtain  B'a  =  2  ma  =  A  —  B,  .-.  ma 
A-B 


30  ARTICLE    181. 

3d. 

From  Geometry  2  m/,2  -f  2  (Xm2  =  0 V  -f .  O7 If  =  (2  0'  m)2 
=  (A  -|-  B)»  =  ~M\ 

.  •.  as  the  square  of  pq  is  equal  to  the  sum  of  the  squares  of  Q'q 
and  O'p,  tlie  triangle  has  a  right  angle  at  0'. 

4th. 

In  the  equation  on  page  187,  A.  M.,  c  =  sec/  = as 

cosj 

will  be  seen  by  an  inspection  of  Fig.  89,  A.  M.  By  construc- 
tion, —  = —  —  =  cos  /  =  cosB'O''/.  This  last  value 

O'    B'        c  r         c 

is  obtained  from  the  figure,  where  it  will  be  seon  that/  =  WO'a. 
Then  as  cos  B'O'/t,  =  -rf-fin  *"  '°^°ws  that  the  angle  formed 

by  joining  B'c/,  is  90°  at  a,  also  that  as  the  angle  at  a  is  90°,  the 
line  <] p  is  parallel  to  A'O',  and  therefore  tangent  to  the  ellipse 
at  B",  the  extremity  of  the  diameter  conjugate  to  0'  A'. 

5th. 

Call  the  angle  p  0'<t,  W.  Draw  perpendiculars  from  a  to 
0'^  and  (Yq,  and  call  their  intersections  with  these  lines,  E  and  F. 

Then  «E  =  ap  sin  /_p  =  A  sin  (f)()  —  B')  =  A  cos  B'. 

«F  =  aq  sin  /_  q  =   B  sin  I.1/. 
0V  =  a'E2  -|-  aF~*  =  A2  cos"  B'  +  B'2  sin'J  B'. 

By  comparing  this  equation  with  equation  13,  Art.  95.  A  M., 
and  the  notes  upon  that  equation,  it  will  be  seen  that  E'  =  /_p 
0'«  =  the  angle  that  a  normal  to  a  tangent  makes  with  the 
major  axis  of  the  ellipse.  Hence  0'^  is  the  direction  of  the 
major  axis,  and  O'q  at  right  angles  with  it,  the  direction  of  the 
minor  axis. 

Equations  6  are  obtained  from  the  equation  of  the  length  of 
any  diameter  of  an  ellipse,  in  terms  of  the  axes  and  the  angle 
made  by  the  diameter  with  one  of  the  axes. 

Let  k  be  the  angle  made  with  the  major  axis,  and  k'  the  angle 
made  with  the  minor  axes. 


r   A*  sin2  k  -|-  B2  cos2  k       r   A2  cos2  kf  -f  B2  sin2  k' 
(Church's  Analytical  Geometry,  page  1 G9.) 


ARTICLE    182. 


31 


The  last  value  in  equation  6,  which  is  the  value  of  sin  V  ob- 
tained by  solving  the  above  equation,  should  be 

A     .  /<?  r*  —  B2 


ARTICLE  182. 

Theorem  I. — Let  O  sl  represent 
any  flexible  and  inextensible  cord, 
fixed  at  sl  and  0  in  the  plane  X  Y, 
and  'acted  upon  by  forces  in  that 
plane. 

Let  p  be  the  force  per  unit  of 
length  of  the  curve,  and  a  and  ft 
the  angles  that  p  makes  with  the 
axes  OX  and  OY. 

Let  T  —  tension  at  any  point, 
making  the  angles  a  and  b  with 
OX,  0  Y.  Let  TO  =  tension  at  O, 
making  angles  «0  and  b0  with  OX, 
and  OY.  Let  s  —  any  length  of 
the  curve  measured  from  0.  Let 
s,  =  whole  length  of  the  curve 
measured  from  0. 

Resolving  all  the  forces  parallel 

/  .  dx 

to  the  axes  I  since  cos  a  =    - — 


and  cos  b 


-  dl  \ 

~  d  s/' 


we  have 


T (-    j    p  .  ds  .  cos  a  —  TO  cos  GO  =  0. 

T  — ^-  4-     /     «  .  cfo  .  cos  /3  —  T0  cos  b0  =  0. 
d  s          «/    s 

Differentiate  equations  1  and  2. 

T  .  d  (—\  4.  —.  d  T  =  -  p  .  ds  .  cos  a. 

\d  s*        d  s 

T  =  —  »  .  ds  .  cos  B. 


0) 

(2) 
(3) 


dy 


Multiply  3  by  —-.  and  4  by  —  ,  and 
ds  ds 


add. 


32  ARTICLES    182-185. 

'^l^-^-p** 


*   *(*)+*.  0  (*')  =  J  i  (*^+*!)=  0. 
ds       \ds/~ds        \dsf        2        \       c/6'-        / 


And 

Then  rf  T  =  —  j>  (cos  a  .  cos  a  -{-  cos  5  .  cos  /3)  c/s  —  —  p  .  di  . 
cos  0.  (5) 

In  which  cos  .  0  =  the  cosine  of  the  angle  between  the  direc- 
tion of  the  force  p,  and  a  tangent  to  the  curve.  For  in  general, 
cos  0  =  cos  (a  —  o)  —  cos  a  cos  a  -j-  sin  a  sin  a  =  cos  a  cos  a 
-f-  cos  ft  .  cos  b. 

If  the  force  p  is  normal  to  the  curve,  then  0  =  90*,  cos  0  =  0, 
and  T  will  be  constant,  since  its  differential  is  0. 

In  this  case,  equations  3  and  4  reduce  to 


T  .  d  =  -  p  .  ds  .  cos  a  ;    T  .  d    j      =  -  p  .  ds  .  cos  ft. 

Square  and  add  the  above  equations. 

T' 


==PP-  (6) 


In  which  p  is  the  radius  of  curvature.     (See  Church's  Cal- 
culus, Art.  102.) 

Equations  2. — See  Church's  Calculus,  Art.  106. 

ARTICLE  183. 

/,  x 2    x 2 

x  d  x  =  w  — — - — % 

/»*!  a.  2    yfl 

and  the  equation  above  8  should  be  w    I      xdx  =  iv  — ' — 

ARTICLE  185. 
The   equation    following    12.      When   i  =  0,   the    fraction 

i  —  cos  .  i  sin  i    . 
^ — r-^-: —   -    is  indeterminate.      Its  value    is    found    bv 


ARTICLES    187-197.  33 

taking  the  third   differential  co-efficient   of  the  numerator  and 
denominator,  and  introducing  the  value  i  =  0.     The  result  is 

-  -  •     Hence  for  i  =  0,  py  =  w  r  im  —  ~\  • 

Equal  ion  13  is  found  in  a  form  more  available  for  practical 
use,  on  page  423  of  Rankine's  Civil  Engineering. 

Equal  ion  16.  —  The  cos/  of  this  equation  should  be  sin/. 

AETICLE  187. 
Equation  4.  —  In  Equation  2  we  have  H0  =  max.  value  of 

P,  .  —  ;  —  max.  value  of  =    T  *    •      Differentiating  both  numer- 
d  x  d  x 


ator  and  denominator  with  respect  to  y  to  find  the  value  of  this 

d.  P, 

dy 

vanishing  fraction,  we  have  H0  =  —  —  -  —  (when  y  —  0).      But 


(IP, 


—  2^Q.     Hence,  H0  =   d-  x    (for  y  =  0). 


Equation  2.     P  sin  0  = 
P 


ARTICLE  191. 
P 


ARTICLE  197. 

Equations  5  and  6. — The  last  term  of  the  denominator  of  the 
last  fraction  of  each  equation  should  be  cos2  0. 

Equation  9. — In  the  equation  cos  2  \L,  =  — — ^, 

P\  —  lh 
substitute  the  values  of  i>^  _/>„  and  p.a  from  Equations  1,  7,  and  8, 

cos2  ()  —  1 
and    reduce  to  the   form  cos  2  ^  =  : — 

sin  (b 


34 

cos  0 

\/cos*  0  — 

AI 
COS2  0 

JTICLES 

But 

197-199. 
cos2  0  — 

1 

(1  _  cos2  0\ 

=  - 

sin  0 

sin  0 

sm  0  -.  — 
sm  0 

sin  0 

sin  $     / 
4/1       ^ 

Vcos'2  0  —  cos'2  0 

</<l 

—  cos'2 

0)-(l- 

COS'2  fo) 

.-.  cos 

sin  0 
2  4,  =  cos  0  V  1 
-i  sin  0 

sin'2  ( 

I 

sin2  0 

-   —  sin  0 
k 

sin  0 

sin  0 
-i  sin  i 

sin-0 
3 

=  cos  0  .  cos  sin      -; —  sin  0  .  sm,  sm      -; — 

sm  0  sin  0 

/          .  -i  sin  e\  1  /          .  -i  sin  0Y 

—  cos  104-  sm      -s I  .-.  ,iL  =  -  I  0  4-  sin     -. I  • 

\     '  sin  07  2  \     '  sm  07 

By  changing  the  construction  of  Fig.  57,  A.  M.,  as  directed 
in  Case  2  of  Art.  1 1 2,  to  adapt  it  to  this  case,  it  will  be  evident 

A 
that  2  x  n'  =  2  ^/,  as  measured  in  that  figure,  is  >  180°  .-.  ^/  is 

>   90°.     Therefore  sin     '  must  be   greater   than    90°,   in 

sm  0 

order  that  Equation  9  may  be  true. 

The  reference  near  the  bottom  of  page  216,  should  be  to 
Problem  IV,  £c. 

ARTICLE  198. 

PROBLEM. — Find  the  pressure  against  a  vertical  wall  12  feet 
high,  sustaining  a  bank  of  earth  which  slopes  backward  from 
the  top  of  the  wall  at  an  angle  of  30°,  the  angle  of  repose  of 
the  earth  being  45°,  and  its  weight  100  Ibs.  per  cubic  foot. 

Ans.  3600    /—       —  -\   =  1671  Ibs.  nearly.    (Length  of  wall 

=  1  foot.) 

ARTICLE  199. 

PROBLEM. — A  prismatic  column  of  solid  masonry  80  feet  high, 
and  weighing  120  pounds  per  cubic  foot,  is  to  be  built  upon  a 
foundation  of  earth  whose  angle  of  repose  is  30°.  What  is 
the  least  depth  below  the  surface  of  the  ground  at  which  the 
foundation  course  should  rest,  if  the  surface  of  the  ground  is 
horizontal,  and  the  earth  weighs  100  Ibs.  per  cubic  foot  ? 

Ans.   ICf  ft.- 


ARTICLES    199-215.  35 

Equation  5. — Draw  a  trapezoid  whose  base  b  is  horizontal,  and 
whose  parallel  sides  are  vertical  and  equal  to  p'  and  w  x,  respect- 
ively. The  ordinates  of  this  trapezoid  will  represent  intensity 
of  pressure  at  any  point,  and  its  area,  the  total  pressure  upon 
the  surface.  Divide  the  trapezoid  into  a  triangle  and  a  rectan- 
gle by  a  line  through  one  extremity  of  the  shorter  vertical,  par- 
allel to  the  base.  The  distance  c  b  is  obtained  by  equating  the 
moment  of  the  whole  trapezoid  about  the  center  of  b,  to  the 
sum  of  the  moments  of  the  above  mentioned  rectangle  and  tri- 
angle. The  moment  of  the  rectangle  =  0  .-.  b  .  - — ~ .  b  c 


2               '    6 
c  —  •     .  , —    • In  this  equation  introduce  the  value  of  p' 

as  given  in  Equation  2  of  this  article,  and  the  resulting  equation 
will  be  the  last  value  of  c  given  in  Equation  5. 

ARTICLE  202. 

Equations  10  and  11. — From  what  precedes  these  equations, 
it  would  seem  that  they  should  be  obtained  by  substituting 

—  and  —  — —  for  -  in  Equations  5  and  7  respectively. 
The  following  are  the  results  obtained  by  such  a  substitution  : 
— ; r-r-  w  t  (xz  -\-  h  x)  and  w  I  (x2  -4-  h  x}. 

'>/t?lll  V  '  '  oil  V  '  / 


ARTICLE  214. 

To  obtain  Equations  4  and  5  from  Equations  1  and  2,  divide 
the  numerators  and  denominators  of  the  second  members  of 
those  equations  by  A/~X~,  and  then  make  x  =  infinity  in  the 
resulting  equations. 

ARTICLE  215. 

Equation  7. — Conceive  of  a  circular  chimney  whose  external 
and  internal  diameters  are  t  and  t  —  2  B  respectively.  The 
outer  circumference  is  TT  t,  which,  multiplied  by  b,  and  placed 
equal  to  the  sectional  area  of  the  masonry,  gives 

rlt  =  ^  -  I  (t  -  2  B)' .-.  1  =  B  (l  -  1)  . 


36  ARTICLES  217-260. 

ARTICLE  217. 

NOTE. — The  arm  of  the  couple  will  "be  easily  obtained  by  draw- 
ing from  D  and  F,  Fig.  101,  A.  M.,  lines  perpendicular  to  H  P; 
from  F  a  line  parallel  to  H  P,  and  a  horizontal  line  through  D. 

The  first  of  these  lines  is  — -  cos  6.     The  remaining  term  (y  -}-  •!•) 
t .  sin  (0  -f-  i)  is  easily  obtained. 

ARTICLE  225. 

"  Find  the  center  of  gravity  of  the  load  between  the  joint  of 
rupture  C,  and  the  crown  A,  and  draw  through  that  center  of 
gravity  a  vertical  line."  See  Fig.  107,  A.  M.  "Then  if  it  be 
possible,  from  one  point  in  that  vertical  line,  to  draw  a  pair  of 
lines,  one  parallel  to  a  tangent  to  the  soffit  at  the  joint  of  rup- 
ture, and  the  other  parallel  to  a  tangent  to  the  soffit  'at  the 
crown,  so  that  the  former  of  these  lines  shall  cut  the  joint  of 
rupture,  and  the  latter  the  keystone,  in  a  pair  of  points  which 
are  both  within  the  middle  third  of  the  arch  ring,  the  stability 
of  the  arch  will  be  secure ;  and  if  the  first  point  be  the  point  of 
rupture,  the  second  will  be  the  center  of  resistance  at  the  crown 
of  the  arch,  and  the  crown  of  the  true  line  of  pressures." 
Rankine's  Civil  Engineering,  page  442.  Let  the  student  make 
the  above  calculation  and  construction  for  an  assumed  circular 
arch. 

ARTICLE  234. 

Example  1. — The  value  of  Px  is  obtained  by  remembering 
that  the  thickness  is  small  compared  with  the  radius  of  curva- 
ture, and  that  the  surface  of  a  zone  is  equal  to  the  circumference 
of  a  great  circle,  multiplied  by  the  altitude  of  the  zone. 

ARTICLE  235. 

Equation  5. — This  equation  will  be  found  in  Chauvenet's 
Trigonometry,  Formula  53,  Page  163,  and  Equation  6  in 
Formula  355,  Page  256. 

ARTICLE  249. 
See  reference  made  in  Article  263,  to  this  Art. 

ARTICLE  260. 

After  CD  in  equations  1,  read  a  x.  At  the  end  of  next  to  the 
last  line  on  page  282,  read,  princiml  elementary  A-A/Y////.S. 


ARTICLE    271.  37 

ARTICLE  271. 
Equation  1.  —  See  Art.  179,  Theorem. 

ARTICLE  273. 

Equation  9.  —  In  Equation  6,  let  p0  =  f,  divide  both  numerator 
and  denominator  of  the  last  value  of  p0  by  r2,  and  solve  the  resul- 

T> 

ting  equation  with  reference  to  -- 

Figure  119.  —  The  general  equation  of  Hyperbolas  of  Higher 
Orders,  is  ym  xn  =  a. 

x  y  =  a  is  the  equation  of  a  hyperbola  of  the  first  order, 
referred  to  its  asymptotes,  x  y1  =.  a  is  the  equation  of  a  hyperbola 
of  the  second  order. 

.-.  xy2  —  a  =  z'  y'2.  (A) 

Take  0  R  as  the  axis  of  Y,  and  a  vertical  through  0,  as  the 
axis  of  X. 

Then  we  have 

x  :  x'  ;  ;  yn  :  y1  from  equation  (A) 

r-A:BB::R«_:r-  I  a*  A.  M. 

:  :  o  K*  :  o  r»  f 

By  comparing  these  proportions  with  equation  A,  we  have 
rA  X  r2  =  EB  X  R2  =  a.  (B) 


The  area  r  ABE 


C*xdy  =   C*  —  dy 

J     r  V     f-  * 


(-1\ 
), 


By  comparing  equation  C  with  the  last  equation  in  case  2,  it 
will  be  seen  that  this  area  represents  case  2,  and  that  a  = 

r  A  x  r\  (E) 

From  the  proportion  C  A  :  D  B  ;  ;  q0  :  ql  ,  we  have  C  A  ql 
=  D  B  </0.     By  introducing  the  values  of  ql  and  q0  ,  as  found  in 


the  equations  at  the  bottom  of  page  292,  A.  M.,  the  equation 


q0  , 
.  M., 


38  ARTICLE    277. 


becomes  -  m     CA  =        -  _  m      DB  =  (RB  -  TO)  C  A 

—  (r~A  —  m)  D~B 

R  B  x  C  A  —  7A  x  D~B  =  TO  (CA  —  DB)  —  m  fA-RB) 
RB  (^A  —  rC)  —  (r^A  .  R~B  -  H3)  =  m  (rA  —  R~B) 
r~C  (iA  -  RB)  =  7/1  (^A  -  RB) 
.-.  m  =  7G. 

Hence  it  will  be  seen  that  r  C  represents  m  of  case  1,  in  the 
solution  of  that  case. 

Since  r  C  is  m,  and  r  A  X  r*  —  «, 
We  have  from  equation  4,  q  =  —  —  m  =  —  _  -  --  r  C. 

But  r~A    :  x'  ;  ;  r'2   :  r2  .-.  ^-  =  *. 


.:  q  =  x'  —  rC  =  the  segment  of  the  ordinate  correspond- 
ing to  y  —  ?y,  between  C  D  and  the  curve  A  B. 
Again,  p  =  —  +  m  =  a/  +  r~C  =  a'  +  rlS. 

=  the  entire  ordinate  from  E  F  to  A  B. 
From  6,  we  have 

Po  +£+  m  =  -TA2H1+  r-O  -  r-A  +  r~E  =  AE. 

ARTICLE  277. 
Differentiate  Equation  ]. 

ws  dx=fds-  ^-dx  =  If.   Then^  =  logeS  +  C. 

W 

When  a;  =  0,  S  =  —  —  ;  see  equation  (1)  A.  M. 

Then  0  =  log,  ^L  +  C  or  C  =  -  loge-^. 

' 


=J!        /S  W        !Lf! 

e  x  =V°r       "7"  e  ' 


ARTICLES  295-298.  39 

AETICLE  295. 


Table. — Bead  m'  =  -,  instead  of  -. 
h  y 

ARTICLE  298. 

ft]  nation  1.  Let  G, ,  G.t,  Gs,  and  G  be  the  centers  of  gravity 
of  A, ,  A2 ,  A3 ,  and  A  respectively. 

Take  moments  about  a  horizontal  line  passing  midway  between 
the  top  of  the  uppar  f_a  ige,  and  the  bottom  of  the.  lower  flange. 

A;  =  A,  (^L±VM.  _  |:)  _  Ai  ^^A+A,  _  ^ 


_h_  h_(lh+l>^  A,  -  (h,+l,3-)  A1-(/^~7<1)  A3 

'  '  Jb  ~  2  2  2  A  '   (  ' 

.Equation  2. 


_  A,  (S,  +  *.,  +  2  /,.)  +  A.  (*,  +  *,) 
~~~~ 


,| 


22  2 

A,  (/,,  +  ^2  +  2  7,s)  4  A3  (A2 


_A2(/,2-f  /.,)  - 
~"~ 


The  moment  of  inertia  of  the  section,  about  its  neutral  axis 
will  be 


-j       -f  A,  X  G~^  +  A2  X  G, 


See  the  end  of  Art.  95,  A.  M. 


40  ARTICLE    298. 


12  '    4A2L 

4  2  A,  A2  A3  (A,  4-  A2  4-  2  A3)  (A,  4-  A3) 
4  A!  A32  (Ax  4  A3)2,  4  A2  A:2  (A,  +  A2  4-  2  A3)2  4-  2  Al  A,  A,, 

(A,  4  A2  4-  2  A3)  (A2  4  A3) 

4  A2  A"  (A2  4-  A3)2,  4  A3  A2  (A2  4  A3)2  -  2  A,  A2  A3  (A2  4-  A:!) 

(A,  -t-  A3)  4-  A,  A2  (A,  4  A3)2J.  (D) 

But  2  Al  A2  A3  (Ax  4-  A2  4  2  A3)  (A,  4  A3)  =  2  AL  A2  A3  (Jtl  4-  A:!)J 

+  2  A,  A2  A3  (A2  T  A:))  (/*!  +  A,)- 
2  A!  A2  A3  (Ax  4  A2  4-  2  A3)  (A2  4-  A3)2  =  2  A,  A2  A3    (A2  -f  A3)2 

'   4-  2  A,  A2  A3  (A2  4  /<8)  (Ai  +  A3). 
.-.  2  At  A2  A3  [(A!  4-  A2  4  2  A3)  ((Aj  4-  A3)  4  (A2  4  A3))  —  (A2  4-  A3) 

=  2  A,  A2  A,  [(X  -f  A,)'  4  (A:  4  A3)  (A2  ,   A3)  4-  (A2  +  /,,)«]. 
=  A,  A2  A3  [(A,  4  A3)2  4  2  (h,  +  A3)  (A,  4  A3)  4-  (A2  +  A,)'-'] 

4-  At  A2  A3  (A,  4-  A3)2  4-  At  A2  A3  (A2  4  A3)2 
=  Aj  A2  A3  (A!  4-  7i,  -f  2  A3)2  4  A,  A2  A3  (Ax  4-  A,)2^  A2  A3 

(A2  4-  /»,)». 
This  reduces  equation  D  to  the  form 


A2  At2  4  A,  A2  As)  4-  (h,  4  A3)2  (A:  A/  4-  A3  A/ 

4-  A,  A2  A,)  4  (A2  4  /.,)'  (A,  A32  4-  A3  A22  4-  A, 

A2  A3)J.  (E) 

A,  A22  4  A2  A2  4  A,  A2  A3  =  (A,  4-  A2  +  A3)  Al  A.2  =  A  A,  Ar 

A,  A,4  4-  A3  A2  4  A!  A2  A3  =  A  A,  A,. 

A2  A32  +  A3  A22  +  A,  A2  A3  =  A  A2  A&. 

These  values  reduce  equation  E  to  the  form 


4  A,  A3  (At  4  A3)2  4-  A2  A3  (A2  4-  A:i)2].         (2) 
Equation  3.  —  See  Art.  294,  A.  M. 

E<]  nation  4.  —  This  equation  can  be  obtained  by  considering  A,, 
//„  and  A3  as  being  so  small  that  they  may  be  left  out  of  consid- 
eration. We  should  then  have  A  =  At  4-  A2  and  A'  =  A3  =  A. 
These  values  in  equations  1  and  2,  give 


ARTICLE    300. 


A1      //  A2  —  h'  A'        hl  /  Aj  +  A2 — A2  +  Al 
/»— —      —%J—      =  TV—      ~A~ 

[  =  _L  (At  A2  (2  //)*)  =-  h  ' '  Al  •  4l . 
4  .A.  A. 


Introduce  these  values  in  equation  3 

/».//'.  A,  A. 

Then  M0  =  mWZ  =  AI  =  -  ;/  Af        -/.  V  A,.  (4) 


ARTICLE  300. 

THE  LAST   OF   EQUATIONS  2.  —  To   obtain   the   last   value   of 
--,  multiply  -T-T-  by  =  —  ,  and  divide  by  its  equivalent       ° 


d  x. 


, 

IX.   TVte  r«/«e  o/  «".     By  making  the  proper  substitutions  in 
Equation  5,  we  have 

n"  c2  =    I"    f  —  -  —  d  x*  =  c    C  loge  -  -  — 

t/Ot/oC   —    X  l/o"c    —   X 

Integrate  by  parts  .    /  u  d  v  =  u  v  —    I  v  d  u  =  the  general 
formula. 

Let  loff    -  =  u  and  d  x  =  d  v. 
&fl    c  —  x 


Then  c 


r\          c  c 

/    log.  --  =  c  x  log.  -- 
J   o     "e    c  —  x  ye   c  —  x 


—  cc 
—  a;  =  z.     Then  d  x  =  —  c?z;  se  =  c  —  2 


rc   x  dx  r°  d  z  (c  —  z) 

I    —  —     /    s —   —  c  loga  2  4- 

•    J   o  c  _  x  «/   o  z 

ge(c-^)4-(c-^)[oC 

C      •    ^a;  = 


=  c  -j  x  logc  c  —  a;  loge  (c  —  x)  +  c  log,  (c  —  z)  -  (c  —  x)  j-  ^ 


42  ARTICLES  307-312. 


=  c   I  x  loge  c  -f  (c  —  x)  loge  (c  -  x)  -  (c  —  x)  j-  o    =c2 
.•.  w"  c2  =  e*  or  n"  =  1. 

X.     Value  of   n"   c2    -     f   f      /  =  c   f    d  x  . 

J     0   J     0        VC2    _     3.4  </     0 


-1    X 

sin     — 
c 


Then  —  =  sin  y  ;  d  x  =  c  cos  y  d  y 

.:  c  J    d  x  sin     -  ^    —  c2     /    T/  cos  y  d  y  —   (by   parts)  c2 

•|  y  sin  y  —     I    sin  y  d  y  (     =  c2    •}  t/  sin  y  +  cos  w  1     =  c2 
(  '-/  )  o  (  )  o 


.    -'    u,  -ix  -  1    u;    i 

sin     -  —  sin  .  sin        --  j-  cos  .  sm        -  r    = 

) 


c      o 


ARTICLE  307. 

Remark. — Near  the  bottom  of  page  334,  read,  m"  is  never  less 
than  |. 

The  last  member  of  Equation  3  should  be  — .; ,  • 

m"  m  I 

Equation  13  should  be  M  —  M,  =  — — - — -        —  M,  = 

«,'(#_  a?)'         3  we2    __    3  w  c2    /I          4  x2 
2  8  ~~8 \3  ~~  ITc2" 

ARTICLE  309. 
Equation  3. — Read  /      y  z  d  y  for    /      y  x  d  y. 

J     0  I/O 

ARTICLE  312. 
Equation  7. — Read  —  for  the  co-efficient  of  the  second  mem- 


ARTICLES    314-318.  43 

//  "I    Q  f\ 

ber.     Also  for  —   —  (below  7),  we  should  have  -^r  =    1Q9 
nearly. 

ARTICLE  314. 

In  the  first  place,  suppose  the  external  load  W1  is  made  up, 
in  part,  by  the  weight  of  the  required  beam.  Then,  having  cal- 
culated the  breadth,  and  found  the  weight  B1  of  the  beam,  it 
follows  that,  if  W1  represents  the  whole  load,  W  —  B1  will 

represent  the  load  exclusive  of  the  beam,  or  ^ —  —  the 

ratio  of  the  gross  load  to  the  load  exclusive  of  the  beam. 

Equation  1. — The  student  needs  to  bear  in  mind  that  &1  is  the 
breadth  to  bear  the  gross  load,  minus  the  weight  of  the  beam, 
and  W  is  the  external  load.  Since  h  is  a  fixed  value  in  this 
case,  we  have  from  M0  =  m  W  I  =  nfb  A2  (see  page  316),  and 
above  ratio,  W  :  W1  —  B1  ;  :  b  :  bl 

•'•  b  =  w'-'u'  <1} 

Equation  '2. — The  depths  being  constant,  the  weights  of  the 
beams  are  proportional  to  their  breadths. 

•  b1  W1  W 

...B^B::^::^^::!:^^,^ 

B1  "W1 
\\n  _  B1 

Equation  3. — The  true  gross  load  will  be 
B1  W  W2 


ARTICLE  318. 
Equation  5. —  From  (4) 
d-x  P  2  dx  .  d*  x  P  /  (I  x 


44  ARTICLES    314-318. 

=  — — -  I  when  c 
c      \ 


dx 


a 
When  x  =  0,  y  =  0  .-.  c  =  0,  and  the  above  equation  becomes 

y          x  y  y 

—  or  —  =  sin  —  .  •.  a;  ==  a  sin  —  • 
a  c  c 


STRENGTH  AND  DEFLECTION 

OF 

HORIZONTAL  BEAMS 


STRENGTH    OF    BEAMS. 


NOTATION. 

I      =  Length  of  beam. 

c      —  ^  the  length  of  a  beam  supported  at  both  ends,   or  the 

length  of  a  beam  fixed  at  one  end,  and  free  at  the 

other. 

b      =  Breadth  of  beam. 
A     =:  Depth  of  beam. 

y     —  Distance  from  neutral  axis  to  outer  fiber  of  beam. 
•to     =  Load  per  unit  of  length  of  beam. 
W  =--  Total  load. 

R     =  Force  at  right  hand  support. 
L     =  Force  at  left  hand  support. 
M    =  Moment  of  external  forces,  at  any  cross-section. 
M0  =  Maximum  value  of  M. 
F     =  Shearing  force  at  any  cross-section. 
F°   =  Maximum  value  of  F. 

I     =  Moment  of  inertia  of  cross-section,  about  its  neutral  axis. 
E    =  Modulus  of  elasticity. 
/     — -  Intensity  of  stress  at  the  outer  fiber. 
yl    =  Maximum  deflection  of  beam. 
m    =  A  constant  factor  depending  upon  the  method  of  loading 

and  supporting. 

n     =  A  constant  factor  depending  upon  the  form  of  the  beam. 
k     =  Coefficient  of  maximum  deflection. 
O    =  The  Origin   of  a  system  of  rectangular  coordinates,   in 

which  the  axis  of  the  undefiected  beam  is  taken  as  the 

axis  of  X. 

The  following  notes  on  the  strength  and  deflection  of  beams, 
contain  the  substance  of  what  is  found  in  the  A.  M.,  from  Arts. 
288  to  308,  inclusive,  and  may  be  substituted  for  those  articles. 
It  is  believed  that,  from  the  simple  demonstration  of  the  fun- 
damental formulae  for  the  restricted  case  of  horizontal  beams 
with  parallel  vertical  loads,  the  average  student  will  proceed  to 
their  application  more  rapidly  and  with  greater  independence 
and  security  than  by  following  the  Applied  Mechanics.  These 
pages  are  therefore  intended  to  be  complete  in  themselves. 

Rankine's  notation  has  been  followed  as  far  as  practicable, 
and  whenever  introduced  the  letters  stand  for  the  same  quanti- 
ties as  in  the  A.  M. 


STRENGTH    OF    BEAMS.  47 

All  the  notation  will  be  found  explained  on  page  46,  or  in 
connection  with  the  subject  where  it  occurs. 

In  all  problems  relating  to  the  strength  of  materials,  two  dis- 
tinct systems  of  forces  must  be  considered. 

1st.  The  external  forces,  consisting  of  the  applied  loads,  and 
the  supporting  forces. 

2d.  The  stresses,  or  forces  acting  amongst  the  fibers  or  par- 
ticles of  the  material. 

As  these  two  systems  of  forces  stand  related  to  each  other  as 
cause  and  effect,  it  follows  that,  so  long  as  equilibrium  exists,  the 
two  systems  must  balance  each  other. 

If  we  conceive  of  a  cross-section  made  at  any  point  of  a  beam, 
we  can  imagine  the  beam  broken  by  the  particles  sliding  upon 
each  other,  along  that  section  (shearing),  or  by  the  pulling  apart 
or  crushing  of  the  fibers  in  a  direction  perpendicular  to  the  sec- 
tion (cross  breaking).  Since  a  part  of  the  fibers  are  extended 
and  part  compressed,  these  stresses,  of  opposite  kinds,  form  a 
moment  about  the  neutral  axis  of  the  section  which  tends  to  pre- 
vent the  beam  from  bending,  and  which  is  equal  to  the  moment 
of  the  external  forces,  or  "bending  moment,"  M. 

To  find  the  value  of  F,  take  the  algebraic  sum  of  all  the  ex- 
ternal forces  acting  upon  the  portion  of  the  beam  included  be- 
tween the  section  and  one  end,  considering  the  upward  or  sup- 
porting forces  as  positive,  and  downward  forces  or  loads  upon 
the  beam  as  negative. 

M  is  found  by  taking  the  moment  of  all  the  external  forces 
acting  on  the  beam  between  the  section  and  one  end,  about  the 
neutral  axis  of  that  section.  We  have  found  it  convenient  in 
most  cases  to  use  right  hand  rotation  as  negative,  and  left  hand 
rotation  as  positive. 

The  maximum  values  (F())  of  F,  and  (M(l)  of  M,  are  generally 
required.  The  former  is  evidently  some  part  of  the  whole  load 
(W),  and  the  latter  some  portion  of  the  product  of  the  load  and 
length  of  the  beam,  and  is  expressed  by  the  equation  M0  — 
mWl 

Beams  of  sufficient  strength  to  resist  the  bending  moment, 
have  usually  ample  strength  to  resist  the  shearing  force  ;  hence 
any  further  consideration  of  it  here,  will  be  omitted.  (See  Art. 
309  A.  M.,  and  Table  TI  in  the  Appendix.) 

To  determine  the  manner  in  which  the  beam  resists  the  bend- 
ing moment  M,  notice  that  when  the  beam  is  slightly  curved,  the 
layers  of  fibers  upon  the  convex  side  are  extended,  while  those 
upon  the  concave  side  are  compressed,  so  that  there  will  be  one 
surface  in  which  the  fibers  are  neither  extended  or  compressed. 


48  STRENGTH    OF    BEAMS. 

This  is  called  the  "  neutral  surface."  It  is  assumed,  (which  is 
near  enough  to  the  truth  for  most  practical  cases,)  that  this  sur- 
iace  passes  through  the  center  of  gravity  of  the  cross-sec- 
tion of  the  beam,  and  contains  the  neutral  axis  of  that  section. 

Within  the  limits  of  proof  stress,  (see  Art.  245  A.  M.,)  the  co- 
efficients of  extension  and  compression  are  exactly  or  nearly 
equal.  From  Hooke's  law  that  the  strains  are  proportional  to 
the  stresses  producing  them,  we  have  zero  as  the  stress  at  the 
neutral  surface,  and  at  any  other  point  it  is  proportioned  to  the 
distance  of  that  point  from  the  neutral  surface. 

SECTION  1.  —  Moment  of  Bending  Stress. 

Let  p  =  the  intensity  of  stress  at  any  distance  y,  fromOZthe 
neutral  axis  of  the  section. 

Let  /  =  intensity  of  stress  at  outer  fiber. 

d  y  .  d  z  =  an  elementary  area. 
p  .  dy  .  dz  —  stress  on  an  elementary  area. 
p  .  y  .  dy  .  dz  =  moment  of  this  stress  about  the  neutral 
axis. 

'  d    .  dz  .  =  the  total  moment. 


/  / 

«/    —     J    ~f 


Since  the  stress  is  zero  at  the  neutral  surface,  and  increases  to 
/  at  the  outer  edge,  and  is  a  uniformly  varying  stress,  we  have 

2>  .:  /  :  :  y  :  ye  w  j»  =  —  y. 

This  value  introduced  in  the  above  integral  gives, 


2/0 

which  is  the  moment  of  bending  stress  at  any  section.  A  few 
values  of  I  are  given  in  a  table  at  the  end  of  Art.  95,  A.  M. 

From  these  values  it  will  be  seen  that  I  =  a  b  A3,  in  which  a 
is  a  constant,  depending  upon  the  form  of  section. 

Also,  that  y0  =  dxh,  where  d  is  a  constant,  depending  upon 
the  position  of  the  neutral  axis  of  the  section. 


.-.  -  -  =  —  —  -  =n  f  b  7t2,  where  n  is  a  constant  and  =  -T- 
y«  d  h  d 

ome 

(B). 


Hence  the  general  formula  for  the  moment  of  bending  stress 
at  any  section  is, 


STRENGTH    OP    BEAMS. 

TABLE  ]. 


49 


VALUES  OF 


Rectangle  b  h,  including  square,    

Hollow  Rectangle,  b  h  on  outside,  and  b  h  on 
inside, 


Circle  and  Ellipse, , 


Hollow  Ellipse,  outside  diameters  b  and  h,  in- 
side Z>y  and  hj, 


32 


tl  - 

V 


Hollow  Circle,  diameters  h  and  h , 


-      1  - 
32 


For  other  forms  of  cross-section,  calculate  n  from  the  formula 
1 


SECTION  2. —  Transverse  Strength    of  Beams    of  Uniform  Section. 

Calculations  upon  the  transverse  strength  of  beams  are  based 
upon  the  principle  of  equality  of  moments.  The  maximum 
moment  of  the  external  forces  must  be  equated  to  the  moment 
of  bending  stress  at  the  cross-section  of  the  beam. 

Thus 

M0or™WJ  )        (    fl 

Mo.  of  external  forces  j  —  ^  ~~^~  —  n  J  °  h  I  /Q\ 

Mo.  of  internal  stress 


PROBLEM  I. —  To  find  the  safe  load  for  a  given  beam. 

The  modulus  of   rupture  of  ash   being   12,000  pounds  per 
square  inch,  find  the   safe  load  that  may  be  uniformly  distrib- 


be   M   =    R  (c  —  x)  —  w  (c  —  x)    (c  ~~  x\    = 


50  STRENGTH    OF    BEAMS. 

uted  over  a  rectangular  ash  beam  whose  cross-section  b  x  h  is 
3"  x  12",  and  length  I  =  20/?.,  when  supported  at  both  ends. 

As  the  modulus  of  rupture  is  usually  given  in  pounds  per 
square  inch  (see  Table  IV  in  the  appendix  of  A.  M.),  the  cross- 
section  and  length  of  the  beam  should  have  their  dimensions 
expressed  in  inches. 

Solution  I  —  20  x  12  =  240" 

The  whole  load,  ~W  =  wl  =  2wc  =  20  x  12  X  *»  =  24G<r. 
The  supporting  forces  R  and  L  will  each  be  we  =  l'2Qw. 

Let  the  origin  0  be  at  the  center  of  the  beam,  and  take  a  sec- 
tion at  a  distance  x  to  the  right  of  0.  Then  the  moment  at  that, 

section   will 

B(.-«  )- 

The   maximum  value  of   M  will   be   obtained   by  making 

^  Or  M,,  =  Re  _  -£.  =  Wff  _  J^.   =    *.  =  1  W  I  = 

30  W. 

By  Art.  247  A.  M.,  it  will  be  seen  that  Rankine  considered 
10  as  the  proper  factor  of  safety  for  timber,  so  that  the  outer 
fiber  should  not  be  under  a  stress  greater  than  12000  -4-  10  = 
1200  Ibs.  per  square  inch. 

.-.  /  =   1200.    I  =    -^-  Substitute  in  the  equation  M0  =  — 
and  we  have 

1200    X 

2 

2880 
W  —  2880  Ibs.,  which  is  rr^  144   Ibs.   per   lineal   foot 

of  beam.     Ans. 

PROBLEM  II.  —  Knowing  the  form  of  cross-section  of  <i  given  beam, 
and  tlit  loa<l,  to  find  its  dimensions. 

If  the  modulus  of  rupture  of  cast  iron  beams  is  40,000  Ibs., 
find  the  cross-section  of  a  rectangular  beam  25  feet  long,  and 


30W  =  -       —  —  —      —  =  200  X  3  x    144. 


NOTE.— The  student  mii^t  continunlly  have  in  mind,  that  all  weights  or  measures 
lust  be  reduced  to  milts  <>r  il<e  KHUH  di-noniinniion 
NOTE.— The  subject  of  lactoi-s  <;!'  .-alety  i*  e^yluined  in  Arts.  -245-247,  A.  M. 


STRENGTH    OF    BEAMS.  51 

b          I 
whose  breadth  and  depth  shall  have  the  ratio  — -  =  - ,  for  a  load 

/I  O 

of  200  Ibs.  per  lineal  foot,  and  a  concentrated  load  P  =  2200 
Ibs.  at  the  center,  the  beam  being  supported  at  both  ends. 

?  —  25  x   12  =  300"  .  c  =  y    =    150"  .     W  =  (25    x    200) 

-f  2200  =  7200  Ibs.   w  =  2.5,  X  ^    =]-K-'"{  Factor  of  safety 
Zo  X  1^  «* 


=  Q}.f=  40000  4-  6  =  —3—;    »  =  g(See  Table  I)  .  MO 

;   R  =  L  =  3600. 

Let  the  origin  O,  be  at  the  center  of  the  beam,  and  take  a 
section  at  a  distance  x  to  the  right  of  0.      Then  we  have 
x  w 

.-.  M0  =  Re  -  Y  c'  =  360°    X  150    -  ^  y  x    1502  =    150 
(3600  —  1250)  =  352500. 

..  352500  =  iv^v'v 


=  —  =  3.28"  nearly. 


PROBLEM  III. 


Find  the  strongest  rectangular  beam  that  can  be  obtained 
from  a  given  cylindrical  beam. 

Let  D  be  the  diameter  of  the  circular  section  of  the  given 
beam.  We  have  from  the  equation  of  the  circle,  D2  =  &2  -\-  h? 
.-.  h?  =  D2  —  V.  Since  for  the  transverse  strength  of  beams 
M0  =  n  /  5  tf,  we  have  M0  =  nf  (D2  b  -  Z>3). 

The  value  of  I  which  makes  Mo  a  maximum  is  found  to  be, 
T 


The  corresponding  value  of  h  is,  7i  —  D 


DEFLKXION    OF    BEAMS. 


SECTION  3. — Equation  of  the  Elastic  Curve  for  Beams. 
\ 


Resume    the 
equation  M  = 

—     Let  s  equal 

2/0 

the  stress  on  a 
fiber  at  a  unit's 
distance  from 
the  neutral  axis. 
Then  s  :  /  ;  ; 

!:,.«=. 


Let  X  be  the 
elongation  of  a 
fiber  at  a  unit's 
distance  from 
the  neutral  axis. 

Then  d  X  will  be  the  elongation  of  a  fiber  whose  original  length 
was  d  x.  3<ufafy&4JL~ 

Let  p  be  the  radius  of  curvature  of  the  neutral  ^axis/  Let  E 
be  the  modulus  of  elasticity,  that  is,  the  force  that  will  elongate 
a  bar  whose  section  is  unity,  to  double  its  original  length,  pro- 
vided the  elasticity  of  the  material  does  not  change. 

From  Hooke's  law,  "As  the  stresses,  so  the  strains,"  we  have 


/I 

j            -CJ. 
d  X 

rfX    BI 

2/0 
From  the  similar  triai 

Y  •  /7/r  •  •  1  .  x  . 

rfo; 

igles  in  the  figure  above,  we 
1 

have 

P 
El 

.   (to 

But  0  = 


(dx*  +  dff  = 
c^a;  c?2y 


DEFLEXION    OF    BEAMS. 


Since  is  the  tangent  of  the  angle  that  a  tangent  line 

to  the  curve  makes  with  the  axis  of  a-,  and  in  beams  is  exceed- 
ingly small  when  compared  with  unity,  it  may  be  neglected. 
Then  we  have 


y  ••-  —  = 


dx* 


dx* 

is  the  equation  sought. 

PROBLEMS. 

1.     Find  the  maximum  deflection  of  a  beam  loaded  uniformly, 
and  supported  at  the  ends. 
Solution.     (Origin  at  center.) 

w 


=  wc(c—  x)  —       (c- 


__ 

we,  .„        w   ,  .  .    ,    w  c3  a;      _ 

l  y  =  -g-  (c—x)-ii  (c  ~  ^  +—3-  +R 

When  x  =  c,  y  =  0  .-.  D  =  —  ^  - 
we,  .  ,        w  .,       w  <?  x       w 


At  the  point  of  maximum  deflection  C—-  =  0,   or  the  curve 

a  x 

is  horizontal.     In  this  case  it  is  obviously  at  the  center,  where 


NOTE.  —  All  the  beams  in  this  and  the  following  section  are  supposed  to  be  of  uni- 
form cross-section,  horizontal  in  position,  and  with  vertically  applied  loads. 


54  DEFLEXION    OF    BEAMS. 

5      W    P 

384~IT  ' 

2.  Find  the  maximum  deflection  of  a  beam  supported  at  the 

1    P  P 

ends,  and  loaded  at  the  center  with  P.     Ans. — — 

48  hi  I 

3.  Find  the  maximum  deflection  of  a  beam  supported  at  the 
ends,  and  loaded  at  the  center  with  P,  and  uniformly  with    w. 

4.  Find  the  maximum  deflection  of  a  beam  fixed  at  one  end, 

I    P  P 
and  loaded  at  the  free  end  with  P.     Ans.  —  — —  . 

5.  Find  the  maximum  deflection  of  a  beam  fixed  at  one  end, 
and  loaded  at  the  free  end  with  P,  and  uniformly  with  w. 

6.  Find  the  values  of  M0  and  yv  for  a  beam  fixed  at  its  ends, 
and  loaded  uniformly  with  w,  and  at  the  center,  with  P.     Find 
also  the  point  of  contra-flexure. 

Let  w  I  =.  W/ 

W 

(from  formula  E.)   where  p  is  the  moment  required  to  hold 
either  end  of  the  beam  horizontal. 


dy  C 


When  x  =  c,  y  =  0  .-.  D  =  0. 
y  is  a  maximum  when  x  =  0. 


NOTE.— "Fixed,"  muaiib  Uxud  m  direction. 


DEFLEXION    OF    BEAMS.  55 

EIv  =  —  4  —  —  —C-  —  ?—  —  —C-=   _/?_^4-wc*\ 
12-6          24          8    ~      6    ~          Y24       24  / 


_  .  __ 


192       384  /  384 

To  find  the  value  of  MQ,  substitute  the  value  of  p,  from  equa- 
tion 2,  in  1 


\\  liich  gives  a  maximum  value  for  M,  where  x  =  c, 
/P  c       w  c2\          /  P  I       w  I-\          /3  P  +  2 

•••M-MT  +  ir)MT+n>-(Hs- 

To  find  the  point  of  contra-flexure,  make  —  -^  =  0,  in  equation 
4,  and  we  have 


=  it  —  + 1 4-  i/I*i±J  PAv:  +  w/> 

\2WT    •  2—  V  12"  W  -  "     * 


=  distance  from  end. 

7.  Find  the  value  of  R:  the  point  of  Max.  deflection  (a^);  the 
Max.  bending  moment;  and  the  point  of  inflection  for  a  beam 
fixed  at  L,  supported  at  R,  and  loaded  uniformly.    * 

R  =  f  w  Z  =  |  W, 

xt  =  .58  I  (nearly.^  from  L, 

Max.  moment  £  W  /, 

Point  of  inflection  at  \  1  from  L. 

8.  Find  the  point  of  contra-flexure,  and  values  of  R,  yv  and 
M,,,  for  a  beam  loaded  at  the  center  with  P,  fixed  at  L,  and  sup- 
ported at  R. 


56  DEFLEXION    OF    BEAMS. 

First  consider  that  part  of  the  curve  included  between  the 
right  hand  support  and  the  center. 


E 


E  I  y  =  —  (2  c  —  x)3  -f  C  a;  -f  D  .......  (3). 

When  a;  =  2  c,  y  =  0  .-.  D  =  _  2  C  c. 

V  B  I  y  «=  -f-  (2  c  -  »)«,—  C  (2  c  —  -^  ......    (4). 

Now  consider  that  part  of  the  beam  included  between  the 
center  and  the  left  hand  support, 


EI 


When  a/  =  0,  -^~  ==  0  .-.  F  =  2  R  c2  ---  —  =  -^-  (4  R  —  P) 


(4E—  P)  .........................    (6.) 

—  ^')s-^  («  —  •')•+  Y  (411  -  P)  j 


(P  —  8  R.) 

...  E  Iy>  =  -?-  (2  c-*')3  _  I-  (o_,')3 

+          (P-8R) 


rf  y  rf  y' 

When  x  =  x'  =  c    y  =  y'  and  —  j  —  =  -^  —  . 

c^  a;  (/a 

Under   these   conditions,  we   have   from  2    and  6 


DEFLEXION    OF    BEAMS.  57 

.-.  C  =  —  (4  R  —  P),  and  equation  4  becomes 

E  I  yM|..  (2  «_*)«_  -^(4R-P)(2C_*)..    (8). 
We   also    obtain   from   the   above   condition   and   7   and   8, 


(P  -  8  R)  ;  0  =  (4  R  -  P)  -f  -  (P  —  8  R)  =  24  R  —  6  P  _|_ 
P  -  8  R  .-.  R  =  j|  P  ..............................   (9). 

To  find  the   point   of    maximum   deflection,    substitute   the 
values  of  R  and  C  in  equation  2,  and  we  have 


The  maximum  deflection  is  where        *      =    0,   hence   to  find 

5  c2 

the  point,  we  have  0  =  —  ^  (2  c  —  x)'2  +  -§- 

.-.  5  (2  c  -  a;)4  =  4  c2 

(2   c  -  x)  =   ±  2  c  jA 

x  =  2  c  —  2  c  4/ _  =±  >|- /I    — i/—    1  r=  distance  from  fixed 
5  \  5   / 

end.     The  amornt  of  maximum  deflection  will  be  found  by  sub- 
stituting this  value  of  a;  in  equation  8,  which  gives  ?/, 


-  1  0  -  i/i  ))) 


iL       1  i/ 

96   ^125  32    K   5   7     E  I 


l.  -n  PJL 

;ij        967    E  I 

li/rpjl. 

48  r   5    El 


58  DEFLEXION    OF    BEAMS. 

Substitute  the  value  of  R  in  equation  5. 
Then  B  I   ~^2  =   A  p  (2  c  _  x'}  -  P  (c  -  x') (11). 

f/2      y' 

To  find  the  point  of  contra-flexure,  make  -r-jjjj  =  0 
Then  —  P  (2  c  -  a/)  -  P  (c  -  x')  =  0 


a/  =  --  -  c  =  —  Z  =  point  of  contra-flexure. 
To  find  M0  from  equation  1  1  we  have 

M  =    T^   (10  c  -  5  aS  \  16c  +  16.^)  =  ^  (11  of  -  6  c) 
Making  x'  •=  0,  we  have 


9.  Find  the  same  quantities  for  a  beam  fixed  at  one  end,  sup- 
ported at  the  other,  and  loaded  uniformly,  and  at  the  center 
with  P. 

10.  Given  a  beam  supported  at  the  ends  and  center,  with  a 
fixed  load  over  its  entire  length,  and  an  additional  load  over 
one  span,  to  find  R,  and  the  point  of  contra-flexure. 

Let  w  be  the  intensity  of  the  fixed  load,  and  w{  that  of  the 
additional  load. 

The  tendency  of  this  load  is  to  curve  the  opposite  end  of  the 
beam  upwards,  so  that  instead  of  requiring  a  support  it  would 
have  to  be  fastened  down  at  that  end. 

Let  the  left  hand  span  be  thus  loaded,  then  R  will  act  down- 
ward. 

Since  the  fixed  load  is  distributed  symmetrically  with  reference 
to  the  center,  in  calculating  for  R,  it  may  be  considered  as  all 
supported  at  the  center. 

Taking  the  origin  of  coordinates  at  L,  and  the  coordinates  of 
the  left  hand  xt  and  ?/,,  and  those  of  the  right  hand  span,  x  and 
y,  we  have  the  following: 
For  the  latter 


DEFLEXION    OF    BEAMS. 


For  the  former 

-r  d-  it  ,   w  x~ 


d  x  2 


These  equations  contain  six  constants,  A,  B,  C,  D,  R,  and  L. 
The  six  equations  required  for  their  determination,  may  be  found 
from  the  following  conditions: 

When  x  =  - ,    y=0.  When  x=l,  yf—  0. 

_  2  _  u  2 

I     d  y        d  y/ 
=  X/=2'    Tx~~d^\ 
d*y      d*  y/ 


Forming    the    equations   and  eliminating  we  have  R  =  — 

16  24  ' 

BP     T         7   jw.l\       7 

±5  =:        :    JU  =:  - 


y-72     „. 

To  determine  the  point  of  contra-flexure,  put  —  ^^==0,  remem- 
bering  that  M  must  include  the  fixed  load. 


x  =  ---  ^  —  =  distance  from  left  end. 


60 


BENDING    MOMENT    AND    DEFLEXION. 

TABLE  2. 


BEAMS  OF  UNIFORM  SECTION. 

YALTES  OF 

Mn  =  mWl. 

Vi 

1. 
j  Loaded  at  the  center  with  W. 
1  Supported  at  the  ends. 
2. 
j  Loaded  uniformly  with  w. 
(  Supported  at  the  ends. 
3. 
j  Loaded  at    free   end  with  W. 
(  Fixed  at  one  end,  free  at  other. 
4. 
j  Loaded  uniformly  with  w. 
\  Fixed  as  in  3. 

5. 
(  Loaded   with   w,    and  at  the 
)      extreme  end  with  P. 
(  Fixed  as  in  3.  (w  l=W). 

6. 
j  Loaded  at  center  with  W. 
(  Fixed  at  the  ends. 

7. 
j  Loaded  uniformly  with  w. 
\  Fixed  at  the  ends. 

8. 
(  Loaded  uniformly  with  w,  and 
-<      at  the  center  with  P. 
(  Fixed  at  the  ends. 

9. 
(  Loaded  uniformly  with  10, 
•j  Fixed    at   one   end  and  sup- 
(      ported  at  the  other. 

10. 
Fixed  as  in  9,  loaded  at  cen- 
ter with  W. 

>.w, 

I   W  P 

;« 

5     W  I3 

384    E  I 

Wl 

1  W  P 

1 

.    1  W  P 

2 

8    E  I' 

2 

£(^+H 

lwi 

1    W  P 

8 

192  E  I 

1.WI 

12 

1     W  /3 
384'  IT' 

(3P-|_2W\ 

(2P+W,)    P 
384       E  I' 

24        / 

!w? 

2.08  W  P 

8 

384     E  I 
nearly. 

3 

i  J/TW  P 

16 

48^   5  EI 

NOTE.  —  To  calculate  the  transverse  strength  of   beams,  use  Ihc  fornuil:).  m\\  I  - 
»/*/*-.    Take  the  value  of  in\\l  from  table  2  and  the  value  of  /<  from  pu^e  49. 


FORMULA    FOR    DESIGNING    BEAMS.  61 

The  preceding  table  gives  the  values  of  Max.  bending  moment, 
and  Max.  deflection  for  the  most  common  and  important  cases 
of  loading  and  supporting.  In  the  column  headed  j\iy  =  m  W  /, 
the  co-efficients  of  W  I  which  are  the  values  of  m  for  the  different 
cases,  are  so  arranged  as  to  be  seen  at  a  glance  and  readily  taken 

W  I3 
out  for  use.     Also  the  co-efficients  of in  the  column  headed 

W  Z! 

y1  are  the  values  of  ~k  in  the  formula  y:  =  k.          ,  or  co-efficients 

Jii  1 

of  Max.  deflection.     (Cases  5  and   8  are  not  included  in  the 
above  remarks  relating  to  the  co-efficients). 

SECTION  4. — Equation  for  designing  a  learn  whose  deflection  shall 
not  exceed  a  given  amount. 

In  the  preceding  problems  we  have  seen  that  the  value  of  y, 

k  W  P 
may  be  expressed  by  the  equation  y\  —  (1). 

-CJ     1 

(2). 

From  formula  6  in  section  2,  we  have  M0  or  m  W  I  =  -  — 
mWly0 

.    .    J_ . 


Substitute  this  value  for  I  in  equation  (1). 

kWPf_kfl2 
' 


In  beams  symmetrical  with  respect  to  the  neutral  axis  of  the 

2k        f        P 

section,  y   =  ±  h;  hence  (3)  becomes  yl  =  —  x  -4-  X  -=-• 

m        xu         n 


A       2k       f        I 

or7=mx.TfxT  (a) 

In  applying  this  formula,  A;  and  m  are  to  be  taken  from  table 
(2),  or  calculated  as  heretofore  explained.  E  and  f  are  found 
in  the  Appendix  to  the  Applied  Mechanics,  tables  1  and  4  re- 
spectively ;  but  the  value  of  /  there  given,  being  the  ultimate 
strength  of  the  materials,  it  must  be  divided  by  a  proper  factor  of 
•yifi'tij  before  being  introduced  into  the  formula. 


G2  li^AIuri    <_>;«'    UMKOiiil    STUENGTH. 

is  the  ratio  of  the  length  of  the  "beam  to  the  greatest 

deflexion  allowable  for  the  particular  case  in  question,  and  is 
assumed  from  experience  or  from  the  necessary  limitations  of 
the  problem. 

PROBLEM. 

Find  the  dimensions  of   the  rectangular  cross-section  of  a 
wrought    iron   beam    20    feet    long,    which   shall   deflect    but 
of  its  length,  for  a  working  load  of  225  Ibs.  per  lineal  foot. 


5 


See  Table  2, 


I  \  See  Table  1, 
o  ) 


B_  27000000. 


Working  strength  of  wrought  iron, 
/  =  9000. 


•'•  i  =  3565!      ?  =  12  X  20   =  240";  W  =  4500  Ibs;  1 
=  1000. 
From  formula  G,  we  have 

5 

h   __    2  jfe        /          I  <  "384" 

T  ~     "^T  '  B    '•  '  ~~T~       *  3000 


50 


720 
From  formula  6,  section  2,  we  have  m  ~W  I  =  nfbh* 

mWl         I   X   45°°    X     24°  81 

:  -  :=    -  ~         = 


_    x    9000    X 


SECTION  5. — Seams  of  uniform  strength. 

CASE  1. — Any  beam  of  uniform  strength,  and  uniform  depth, 
not  fixed  at  both  ends. 

From  the  equation  of  the  elastic  curve,  formula  E,  section  3. 
we  have 


BEAMS    OF    UNIFORM    STRENGTH. 

M 


=  *.-.   E 


d  x>  d  x*     '          I 

M  = or  -:-  =  —  from  formula  6,  section  2. 

2/0  2/o 

Since  the  depth  is  constant,  y0  is  constant  .-.   equation  (I) 
becomes 

cPy         M         / 

E  -~-  =  -y-  =  ^—  =  a  constant (2). 

a  x-  L          y0 

j  -,,  f 

=   -^  x  +  G. 

Take  the  origin  so  that  when  x  =  0,  — ^ —  ^  0,  (generally  at 
the  center  or  end  of  beam)  then  C  =  0 

2  _i_  D.     When  x  =  0,  y  =  0  .-.  D  =  0. 


.-.  E  y   =   -~  —  x2;    y  is  a  maximum  when  x  •=.  c 

w  V    ?/ 

«  yo 

.-.  y.   =    —  ^=  —    =  maximum  deflection,     c  =  ?  or  -  I  ac- 

2-^2/o  2 

cording  to  the  method  of  supporting. 

E  I 

In  section  3,  we  had  the  radius  of  curvature  =  p  =  —  rj— 

As  E  is  a  constant,  and  is  constant  as  above  shown  for 

this  case,  then  p  is  constant,  or  the  curve  is  the  arc  of  a  circle. 

CASE  2.  —  Beam  of  uniform  strength,  uniform  breadth,  built 
in  at  one  end.  free  at  the  other,  and  loaded  uniformly. 

fP    U  f 

As  before,  we  have  E  T-T-  =  ~  — 
d  x*  2/o 

Let  yz  =  maximum  half  depth  of  beam,  and  liz  its  depth, 
Then  M0  =        -  —  nfb  h\ 


(c 

v  #y 

•    &    ~~7  —  : 
ax- 


#  V_       .     Vl_          _!_         J_  /     *'      \ 

—  xf  A*  y0*  y0    "    yz    V  c  —  x  / 


64  BEAMS    OF    UNIFORM    STRENGTH. 

E  --—-    = ^— ,  loge  (c  —  x)  -f-  C 

When  x  =   0  -~-   —   0   .-.  C   =  -^— -  loge  c 

.-.  E  y  =  —     -    /  loge  (  —      —Jdx.      Integrate  by  parts. 

fc(  /       c      \  r  x  dx 

E  v  = 1    &  1(>ge  I  —        —  I   —     /  ~ 

?/z      (  \c  —  x  /          t/     c  —  x 

Make  c  —  x  =  z  and  integrate. 


When  x  =  0,  y  =  0  .-.  F  =  —  (  -  c  loge  c  +  c) 


(c  -  x)  -  c  loge  c  + 

=  -4~    (  lo&<  c  X  (x  ~  c)  +  loge  (c  -  x)  X  (c  —  a:)  +  x  j- 
y  is  a  maximum  when  x  =  <= 

,,  =  _/^  =  _/|_ 

CASE  3. — A  beam   of   uniform   strength,   uniform  breadth, 
built  in  at  one  end,  and  loaded  at  the  other. 

2/1   ==  TEl/T 

CASE  4. — A  beam  of  uniform  strength  and  breadth,  loaded 
at  the  middle  and  supported  at  the  ends. 

y   *2££ 
"'   ~   3  E  y, 

CASE  5. — A  beam  of  uniform  strength  and  breadth,  loaded 
uniformly,  and  supported  at  the  ends. 

E~7/ 

NOTE.— In  case  5  integrate  the  expression  sin~   —  dx  by  making  ein~   -  =  z,or  ^ 
--  sin  z,  and  it  becomes  cz  coszdz.    Integrate  by  parts  and  substitute  the  value  of  z. 


BEAMS    OF    UNIFORM    STKENGTH.  65 

CASE  6.  —  A  beam  of  uniform  strength,  depth,  and  load,  fixed 
at  its  ends. 

For  the  plan  and  elevation  of  such  a  beam,  see  Figs.  144 
and  143  of  A.  M. 

Contra-flcxure. 

As  the  beam  is  of  uniform  strength  and  depth,  the  radius  of 
curvature  will  everywhere  be  the  same.  (See  Case  1.)  And  as 
it  is  fixed  at  C  and  A,  the  point  of  contra-flexure  must  be,  half 
way  between  A  and  C.  (Fig.  143  A.  M.) 

,CB  =     =       .     '  (1.) 


Maximum   deflection. 

It  is  obvious  that  the  portion  BAB  is  in  the  condition  of  a 
uniformly  loaded  beam,  supported  at  its  ends,  whose  length  is 
c  —  ±  L  Hence,  as  the  beam  is  of  uniform  strength  and  depth, 
its  deflection  will  be  found  by  Case  1. 


/(I)' 


/V  f  I* 

y,'  =  -------  =  --—  —  =  -^-=  —  =  vertical  distance  between 

•2^y0         8E.y0       32Ei/0 

A  and  B. 

By  the  geometry  of  the  figure,  it  will  be  seen  that  the  verti- 
tical  distance  between  C  and  A,  is  twice  that  between  A  and  B. 

'"'  y/  =  ~  ' 


Moment  of  flexure  at  A. 

Since  the  portion  B  A  B  is  similar  to  a  uniformly  loaded  beam 
supported  at  the  ends,  the  supporting  forces  are  —  ,  and  the  mo- 

ment anywhere  between  B  and  B  is  M  =  —  I  ---  x    \  --  — 


(I-*)' 


The  moment  at  A  is  given  by  making  x  =  0, 
vc1  _  We        W  I 


66  BEAMS    OF    UNIFORM    STRENGTH. 

Moment  at  C. 

From  B  to  C  the  beam  is  in  the  condition  of  one  fixed  at  one 
end,  loaded  at  the  other  with  --,  and  uniformly  with  w.  Hence 

the  greatest  moment  of  flexure  is  at  C,  where  MC  =  I  ~r  -  1 

c   ,    /  "'  ^  \  e  _  3  ?/;  c2  _  3  W  c  _  3  W  / 
2  "'"  \   2  7  4  ~  ~~8"       ~T6~~  :   :  ~32~" 

Comparison  of  widths  at  A  and  C. 

From  equations  3  and  4,  we  see  that  the  moment  at  C,  is  three 
times  its  value  A.  Hence,  as  the  depth  is  constant,  the 
breadth  at  the  end  must  theoretically  be  three  times  the  center 
breadth. 

See  theoretical  plan  of  the  beam,  fig.  144  A.  M. 

Moment  of  flexure  anywhere. 
For  the  general  expression  for  the  moment  of  flexure,  we  have 


where  p  is  the  moment  required  to  hold  the   ends  horizontal; 

3  AV  c       3  W  I 

hence  it  is  equal  to  the  moment  at  U  =  -  =  -- 

16  32 

see  equation  4.     .-.  M,   =  w  c(c  —  x)  --  —  (c  —  xf  -    —  —  — 


16 


W  /  /••<          .,  V       3  W  I 

ssTi\"*).—§r 


=  the  general  equation  of  the  moment  of 
flexure  for  any  point  in  the  beam. 


ARTICLE  365. 

Equation  1.  —  The  following  method  of  deducing  the  formula 
for  the  centrifugal  force  of  a  unit  of  mass  leads  to  Equation  1, 
and  is  a  good  exercise  for  the  student  in  the  application  of  some 
fundamental  principles  of  dynamics  : 

Let  R  —  the  resultant  accelerating  force  acting  on  a  unit  of 
mass.  Let  the  path  of  this  particle  be  referred  to  three  rec- 
tangular axes  X,  Y,  and  Z. 

Then  since  the  force  is  measured  by  the  product  of  mass  and 

acceleration,  and  the  mass  =  1,  we  have 

—  7—  -  =  component  of  R  along  X, 

-y-f-  =  component  of  R  along  Y, 

<P  z 
,  2-  =  component  of  R  along  Z, 

x 


/  d1  i,  \2  /  d1  z  \2  /  &  s  \2 

+  hf)  +  for)      for) 

d?  s  \2  /  d?  s  \2 

-j-^-J    (subtracting  and  adding  \~^-\  )• 

The  square  of  the  radius  of  curvature  of  any  curve  is 

(See 


Church's  Calculus,  page  146,  Equation  2.) 

Substituting  the  value  of   the  denominator   of   the  second 
member  of  this  equation  for  the  same  quantity  in  the  value  of 


()' 


I  ,  2  I  •  Since  R2  =  the  sum  of  the  squares  of  two  quanti- 
ties, they  must  be  the  rectangular  components  of  R;  and  as 
one  of  these,  (  —p  j  ^  *^e  force  in  the  direction  of  the  mo- 


68 

tion   at  any  instant,   the    other  I  —  I  must  be  the  deviating 

\  p  7 

force  •=.  centrifugal  force. 

ARTICLE  390. 

Equations  1  and  2. — Since  the  radius  vector  is  normal  to  the 
curve,  it  coincides  in  direction  with  the  radius  of  curvature, 
and  both  are  shifted  through  the  same  angle  d  i  in  the  time  d  t. 
The  arc  described  by  the  point  A,  Fig.  187,  A.  M.,  is  measured 
by  p  d  i  ;  or  since  c  r  is  the  velocity  of  A,  by  c  r  d  t.  Hence, 
Equation  (1)  p  d  i  =  c  r  d  t. 

The  distance  through  which  the  end  T  of  the  radius  vector  is 
carried  in  the  time  d  t,  owing  to  the  revolution  about  C,  is  b  r2  d  t, 
measured  along  the  arc,  whose  center  is  at  C.  Its  projection  on 
a  perpendicular  to  A  T  is  b  r2  d  t  cos  6,  which  is  the  arc  subtend- 
ing the  angle  through  which  r  has  been  shifted  negatively. 
This  arc  divided  by  r  gives  the  angle 

br,2dtcos()     .         crdt       br^dtcosQ  /          br^cosd\ 
— ;  a  t  == —  =  a  £  I  c  — 


(0  r2  cos  6\ 
C — f 


ARTICLE  402. 


Read  -  instead  of  n,  in  the  paragraph  preceding  Eq.  (6). 

ARTICLE  421. 

d  .  Q  p         d  .  A  u  p 

Equation  2. — Read  — j-=-*-  =  ? • 

d  s  ds 

ARTICLE  422. 
Equation  (1). — The  last  member  should  read  ^— 

ARTICLE  423. 
Equation  2. — The   second   parenthesis   of   the  first  member 

/dp  dn  dp  d  p  \ 

should  be  I  u  -=-*-  +  v  -=-!»•  +  w  —.  -  I. 

\,  -  -d  x  dy  d  z  d  t  / 

ARTICLE  443. 
The  eccentricity  of  an  ellipse  is  the  distance  from  one  focus  to 

the  center,  divided  by  the  transverse  axis,  or         ~ . 


ARTICLES  444-462.  69 

ARTICLE  444. 

X  Y  =  V(G  H*  +  X  W2).  A  section  of  the  hyperbole-id 
through  X  Y,  and  perpendicular  to  A  B,  Fig.  195,  A.  M.,  would 
be  a  circle  of  radius  X  Y,  and  the  projection  of  the  generating 
line  in  the  position  F  E,  on  the  plane  of  this  section  would  be  a 
vertical  line  at  a  distance  G  H  from  the  center  of  the  section. 
The  point  W  is  where  this  line  cuts  the  circle,  and  its  distance 
from  the  axis  of  the  hyperboloid  is  evidently 

X  Y  =  (G  H2  -f-  X  W*)*.     Let  the  student  draw  the  section  as 
above  directed. 

ARTICLE  450. 

Equation  (1).  The  velocity  of  Pj  in  a  plane  perpendicular  to 
C,  is  «,  C,  PI.  Since  the  line  of  action  makes  the  angle  *  with 
the  axis  c,,  it  makes  (90°  —  «',)  with  the  direction  of  the  motion 
of  Pj.  The  component  of  the  velocity  of  P,  along  the  line  of 
action,  will,  therefore,  be  al  C,  P!  cos  (90°  —  «',)  =  al  C,  P,  sin 
t\.  By  the  same  reasoning  the  velocity  of  P2  along  the  line  of 
action  is  a2  C2  P2  sin  iv  and  these  are  equal  by  the  principle  stated 
in  this  article.  Hence  Equation  (1). 

• 
ARTICLE  458. 

Equation  (1).  r  is  the  length  of  cord  unwound  from  the  circle 
whose  radius  is  C,  P,  (Fig.  198),  while  a  distance  q  on  the  circle 
whose  radius  is  Cl  I  passes  the  point  I.  These  arcs  are  propor- 
tional to  the  radii  of  their  respective  circles,  or 

=  sin  Q, 


ARTICLE  462. 

Equation  (3).  Substituting  values  from  Equations  (1)  and  (2) 
of  this  article  in  the  equation  s  =  2  I—  -j-    —1     /      rdqwe 

have  s  =  4  r0  I  —  +  —  I    /  sin  d  q.     Multiply 

and  divide  the  second  member  by  2  r,,,  and  we  have  *•  =  8  /•  * 


70  ARTICLES  463-483. 

77+7.)  /T(* ; '}  sin  -^  d~^  =  **  VTT+T:, 

(  -  cos  (^    -  o)  +  1)  =  8  rl  (i  +  ij  (l  _  Sin  0). 

ARTICLE  463. 
See  Willis'  Principles  of  Mechanism,  page  137. 

ARTICLE  482. 
The  degree  of  approximation  attained  by  putting 

Vc*  —  (»•,  —  r.2y  —  c  —    •^L——2—  is   seen   by   squaring    both 

members  of  the  inequality.  The  term  '  '  is  neglected  in 
the  second  member. 

ARTICLE  483. 

Equation  (1). — The  first  value  of  y  —  r0  —      '  ~~^    2   is  obvi- 
ous from  an  inspection  of  the  figure  of  the  conoid.     To  obtain 

?•     _l_    •)'  (•/•     <)•  \2 

the  equation  r0  —  —       -  =  — --r-5 —      ^  is  to  ^e  considered 

Z  A    IT    C 

that  the  length  of  belt  for  two  equal  pulleys  of  radius  »•„  is  L  — 
2  c  -(-  2  TT  >•,„  .which,  placed  equal  to  the  approximate  value  of 
L  from  Equation  (3  A),  Article  482,  gives  2  c  -j-  2  TT  rfl  =  2  c 


which  is  also  Equation  (2)  of  this  article.  The  manner  of  using 
this  equation  for  designing  a  pair  of  stepped  cones  in  which 
two  opposite  pulleys  are  of  equal  radii  is  explained  in  the  A.  M. 

If  it  is  required  to  complete  a  pair  of  cones  when  the  radii  of 
a  pair  of  opposite  pulleys  are  given,  the  value  of  rQ  may  first  be 
found  by  substituting  the  given  values  for  rt  and  r2,  in  equa- 
tion (2). 

Then  replacing  rQ  by  this  value,  the  radius  of  one  of  the  re- 
quired pulleys  may  be  assumed  and  the  radius  of  its  opposite 
pulley  calculated  by  solving  the  resulting  quadratic  equation. 

PROBLEM. — In  a  pair  of  stepped  cones  let  the  radii  of  the  end 
pulleys  opposite  each  other  be  r,  =•  24",  r.,  =  6",  and  let  the 
distance  of  their  centers  be  c.  =  48"  required  the  radii  of  the 


AKTICLES    488-191.  71 

adjacent  pair  of  pulleys  on  the  stepped  cones.  From  (2)  we  have 
2r0  =  (r1+r2)+(/'1  ~  '^  =  30  -f-  ^lj\a,,  fr  =  3 }  nearly) 

—  32.15  nearly. 

7T  C 

Let  r,  and  r.2  now  represent  respectively  the  radii  of  the 
greater  and  smaller  of  the  pulleys  to  be  next  calculated. 

We  may  now  fix  the  value  of  r,  and  calculate  r.,.  It  will  how- 
ever generally  be  easier  to  find  r.t  by  trial  than  to  solve  the 
quadratic  equation. 

Let  r,  be  fixed  at  20",  and  for  the  first  trial  let  us  suppose  r, 
to  be  equal  to  11.8".  Then  i\  -j-  ra  =  31.8,  i\  —  r,  =  8.2",  and 

we  have  from  the  formula  31.8   =  32.15  -   L-L-Ll—  —   31.7 

nearly. 

Again  let  r2  =  11.6"  for  a  second  trial,  then  31.6  =32.15 

—  .468  =  31.68.     Thus  we  see  that  the  true  value  of  r,  lies  be- 
tween   11.6"  and   11.8";  r.,  =  11.7"  very  nearly  satisfies  the 
equation. 

ARTICLE  488. 

In  Fig.  218  let  C2  T,  =±  r,  T,  T,  =  /,  the  angle  T,  C2  T,  =  «, 
and  the  angle  T2  Tl  C2  =  /3.  Then  C2  A  =  (/•  cos  a  -f  I  cos  /3) 
tan  (3. 

Placing  the  above  value  of  C,,  A  in  equation  (2),  and  elimina- 
ting p  by  means  of  the  equation,  I  Gin  />  =  r  sin  a,  we  have  rt 

^  cos  a 

?-2  sin  «  <|  —  ^ -|-  1  j>  from  which  the  velocity  of 

j/l-fsin*a 
I  *»      • 

the  piston  of  an  engine  may  be  calculated  for  any  given  crank 
angle;  for  by  knowing  the  number  of  revolutions,  the  value  of 
r2  is  easily  found. 

ARTICLE  491. 

Equations  (2)  and  (1). — Let  the  axis  C2  be  perpendicular  to  the 
plane  of  the  paper  and  let  the  plane  of  the  axes  of  the  shafts  be 
vertical.  The  projection  on  the  paper,  of  the  path  of  F2  will  be 
the  circle  F,  F2  B  D,  and  the  projection  of  the  path  of  Fl  will  be 
the  ellipse  F,  H  B  E  of  the  accompanying  figure. 


Suppose  F2  to  move  to  the  position  F/,  Fj  will  in  the  same 
time  move  to  F/.  The  angle  F2  O  F./  is  02.  The  angle  FT  0  F/ 
which  is  also  equal  to  fa,  is  the  projection  of  the  angle  de- 
scribed by  0  Ft  in  the  plane  of  its  path.  The  true  magnitude 
of  this  angle  is  F,  0  A,  (found  by  revolving  the  plane  of  the 
ellipse  in  which  the  angle  lies,  about  F}  B  until  the  ellipse  coin- 
cides with  the  circle,  when  F,  will  fall  at  A  in  the  vertical  line 
M  A.)  The  angle  P.,  0  A  is  fa.  (See  the  A.  M.) 

Then  0  M  cot  0,  =  M  A.         O  M  tan  02  =  M  F/. 

tan  02         M  F/       O  H 

.-. —  = = =  cos  i  or  tan  0,  tan  0.,  =  cos  i. 

cot  fa          MA        0  F, 

Differentiating  this  equation  we  have 
d  fa        tan  0.2  cos2  fa         tan  fa  (I  -f-  tan'2  0,) 
d  fa        tan  fa  cos'2  fa         tan  fa  (1  -(-  tan'2  02) 

cot^(1+c^-07) 


tan 


COt   0j 


tan  02  =  - 


tan  0.,  -(-  cot  02 
cos  i 


•  cot  0i 
tan  02  4-  cot  02 


Eliminate  fa  by  Equation  (1), 


tan 


+ 


a,  tan2  0,  -f-  cos2  i 

"When  F,  is  in  the  plane  of  the  axes  fa  =  0,  tan  fa  =  0  and 

^1  _ _  which  is  the  first  of  Equations  (1)  A.  M.     In  a  simi- 


ARTICLE    507. 


73 


lar  manner  by  eliminating  p,,  the  second  of  Equations  (1)  may 

be    obtained.      When    —  = p-r—  =  1    or   the 

«!  tan'2  (f>l  -{-  cos-  i 

shafts  have  equal  angular  velocities,  we  find  tan'2  ^  =  cos  i;  <f»l 
—  tan-1.  Vcos  i.  (See  "Willis'  Principles  of  Mechanism,  Article 
513.) 

ARTICLE  507. 

Equations  (3). — The  square  of  the  perpendicular,  drawn  from 
any  point  in  the  circumference  of  a  circle  to  a  diameter,  is  equal 
to  the  product  of  the  segments  into  which  the  foot  of  the  per- 
pendicular divides  the  diameter  .-.  ( --  \  =  T  V  (2  r  —  T  V) 
=  T  V  .  2  C  M. 

An  exact  parallel  motion,  accomplished  entirely  by  means  of 
link  work  (without  any  sliding  piece,  as  in  Watts'  Exact  Parallel 

Y 


Motion,  Article  507,  A.  M.),  has  recently  been  discovered  by 
M.  Peaucillier,  an  outline  of  which  is  here  given. 


74  ARTICLE    507. 

In  the  accompanying  figure,  0  F  E  G  is  an  equilateral  paral- 
lelogram of  bars,  jointed  at  each  of  its  corners.  A  and  C  are 
fixed  points;  A  F  and  A  G  are  links  jointed  at  their  extremi- 
ties. C  E  is  also  a  link  jointed  to  the  corner  (E)  of  the  paralel- 
ogram,  and  capable  of  turning  about  C.  When  the  point  0  is 
moved  in  the  plane  of  the  figure,  the  points  F  and  G  must 
describe  arcs  of  circles  about  A.  E  must  describe  an  arc 
about  C. 

When  this  combination  of  links  is  made  to  oscillate  about  the 
fixed  points  A  and  C  : 

1st.  If  C  is  midway  between  A  and  E,  the  point  0  will  be 
guided  in  a  straight  line  perpendicular  to  A  O. 

2d.  If  C  is  in  any  other  position  on  the  lino  A  E,  the  point 
O  will  move  in  the  arc  of  a  circle. 

To  prove  these  propositions,  and  also  ta  find  exact  relations 
between  the  radius  of  the  arc  described  by  ( ).  and  the  position 
of  C  on  the  line  A  E,  is  the  object  of  the  following  analytical 
investigation  : 

Suppose  the  instrument  to  be  swung  aside  until  the  axis  A  0 
takes  the  position  A  B,  making  an  angle  0  with  its  neutral  posi- 
tion. E  will  be  at  E'  on  this  line,  and  also  on  the  circle  whose 
center  is  at  C.  The  parallelogram  will  be  partially  closed  as 
represented  by  the  dotted  lines  in  the  figure.  O  will  have 
moved  to  some  point  B  on  the  new  position  of  the  axis,  and  it 
is  required  to  find  the  locus  of  this  point. 

Let  O  be  the  origin,  0  X  the  horizontal,  and  0  Y  the  vertical 
axis.  Then  0  M  =  x,  M  B  =  y. 

Let  A  F  =  A  G  =  /;  F  E  =  E  G  =  s-  A  E'  =  c;  A  0  =  A •; 

C  E  =  r ;  C  A  =  d,  and  the  angle  F'  E'  B  =  F'  B  E'  =  ft. 

Then  x  =  (k  —  y}  tan  0  (1) 

y  =  k  —  (c  -f  2  s  cos  ft)  cos  0  (2) 

P  =  c2  -f  s2  +.tc»  cos  ft  (3) 

A  E'  —  c  =  d  cos  0  +  V*1"  —  rf'2  sin*  0  (4) 

Combining  2  and  3  we  have 

(/•-  _  c2  —  s-\                          P  —  s2 
c -| Icos0  =  & cos0         (5) 

c  f  c 

Combining  this  with  (4) 

(?   -   -S2)  COS0 


d  COS  0    -4-     yV  _   (P    gin* 

Putting  0  in  terms  of  tan  0  gives 


ARTICLE    515.  75 

P    —    fS2 


l   -f  tan2  0 

y  =  k  --  -  --  — 


,    i/,.,  _      <(*  tan2  0 
-j_  tan2  0  1  +  tan2  0 

P  —  .v2 


d  -f-  Vr-  4-  (/-2  —  (P)  tan2  0 

Substituting  0  from  (1),  transposing  &  and  changing  the  signs 
of  both  members 

(£  _  y)  (P  —  s*) 


(i  (/c  -  y)  +  V»™  (&  -  y)*  +  **  (''•*  -  c?2) 
Dropping  the  factor  (&  —  y),  clearing  of  radicals,  and  placing 
(P  —  .s2)  =  H,  we  have  (r2  —  rf2)  (A  -  y)2  -f  (r2  —  <P)  x2  =  n2  — 
2nd  (k  -  y)  (7)  (r2  -  </-)  7r  -  2  /,-  (^  —  d*)  y  -f  (»•*  -  a2)  2/2  + 
(r2  —  cP)  .c*  —  n1-  —  1  n  d  k  -\-  1  n  d  y,  which  is  apparently 
the  equation  of  an  ellipse  (since  b-  —  4  a  c  <  0.  See  Church's 
Analytical  Geometry,  page  207). 

Since  the  center  of  this  curve  is  evidently  on  the  line  0  A 
produced,  which  contains  one  of  the  principal  axes,  let  us 
transfer  the  origin  of  coordinates  to  some  point  on  this  line  at  a 
distance  a  from  O.  The  formuke  for  such  transformation  are 
x  =  x';  y  =  y'  -\-  a.  Then  from  (7)  (r2  —  d2)  (k  —  a  —  y')*  -f 
(;.»  _  d')  d*  =  n2  -  In  d  (k  —  a  —  y'). 

Let  k  —  a  =  b,  and  drop  the  primes;  (r2  —  d*)  x2  -f  (r2  —  d2) 
/  =  n*  —  2  n  d  b  —  (r2  -  rf*)  V  +  •(  2  n  d  -f  2  (r2  —  d2)  b  }.  y. 
Jf  the  origin  is  at  the  center  of  the  curve,  the  term  containing  y 

must  disappear  or  2  n  d  -f-  2  (r2  —  d2)  b  =  0;  b  =  -5— — -^  • 

7i  d 

Replacing  b  by  its  value,  we  have  k  —  a  =  —  — -^ 

nd 


V  -  d* 
Since  the  co-efficients  of  x*  and  y2  are  equal,  the  ellipse  has 

equal  axis, 'or  is  a  circle  of  radius  a  =  k  +  — -^-      "When  d 

=  r,  a  =  ex,  or  the  locus  is  a  straight  line  Q  .  E  .  D  . 

ARTICLE  515. 

PROBLEM. — Find  the  formula  for  the  work  performed  upon 
the  piston  of  a  steam  engine  during  one  stroke,  the  length  of 


76  ARTICLE    531. 

stroke  being  /,  the  area  of  the  piston  a,  the  initial  pressure  of 
the  steam  j)l}  the  steam  being  cut  off  at  a  distance  c  from  the 
beginning  of  the  stroke. 

Ans.  a  pl  c  1  1  -j-  loge  —  I  • 

ARTICLE  531. 

Equation  (1).  —  Let  g'  =  the  force  of  gravity  at  any  point  on 
the  earth's  surface  at  the  level  of  the  sea. 

Let  T  =  time  of  one  revolution  of  the  earth  on  its  axis. 

Let  n  =  number  of  revolutions  in  a  unit  of  time. 

Then  n  T  =  1. 

The  centrifugal  force  of  a  unit  of  mass  at  the  level  of  the  sea 
and  latitude  X  is 

v2  (2  ,r  .  R  cos  X  .  «)2       4  7T2  R 

F  —  __  —  v      _  '_  =  _  _  cos  X 
R  cos  X  R  cos  X  T2 

The  component  of  F  acting  opposite  to  gravity  is  f  =  F  cos  X 

4  7T2    R  2    7T2    R 


—  cos 


Gravity  being  diminished  by  this  amount,  if  we  subtract  it 
from  G  the  real  attraction  of  the  earth,  we  shall  have 


The  value  of  ./for  latitude  45°  has  been  found  by  experiment. 
Denote  this  value  by  g^ 

Then  from  the  formula  (making  X  =  45°) 

2    7T2  R  2    7T2  R 


•=         --     —  cos 


For  a  point  at  a  distance  h- above  the  earth's  surface  (h  being 
small  compared  with  R)  we  have  from  Newton's  law  of  gravity 

(-V 

•••9=y''irLh-*  =  9'w-™ 


Performing  the  division  to  two  places 

g   == 


ARTICLES  537-557.  77 

ARTICLE  537. 

Equation  (2) — Let  a  be  the  angle  which  a  tangent  to  the  curve 
makes  at  any  point. 

W  W 

Then  Q  =Wcos  a  =      . 


J~       (dz? ' 
+ 


Since  the  horizontal  component  of  the  velocity  of  the  projec- 
tile is  constant  we  have 

v0  cos  0 

V0  COS  0  =  v  COS  a.       .:  COS  a  =  — • 

v 

v0  cos  0 


.-.  Q  =  W. 


ARTICLE  545. 

Equation  (1) — The  component  of  g  urging  the  pendulum 
towards  its  lowest  point  is  g  sin  6  in  which  Q  is  the  inclination 
of  the  curve  to- the  horizon.  To  show  that  in  the  cycloidal  pen- 
dulum this  force  is  proportional  to  the  length  of  the  arc  between 
any  position  of  the  pendulum  and  its  lowest  position,  we  con- 
sider that  the  length  of  any  curve  is  s  =  /  d  s  =  I  p  d  Q. 
From  Art.  390  A.  M.,  we  have  for  the  cycloid  p  =  4  r2  cos  6. 
.-.  s  =  4  r2  /  cos  0  d  Q  =  4  r2  sin  0  which  varies  as  sin  0. 

*/     o 

Hence  the  vibrations  of  a  cycloidal  pendulum  are  isochronous. 

ARTICLE  552. 

PROBLEM. — A  body  weighing  100  Ibs.  starts  from  a  point  48 
feet  above  the  earth's  surface.  After  falling  freely  16  feet  it 
encounters  a  constant  resistance  equal  to  f  gravity.  Required 
its  potential  energy  before  starting  ;  its  actual  energy,  when  at  a 
distance  of  16  feet  from  the  earth  ;  its  potential  energy  at  the 
same  point ;  the  sum  of  its  actual  and  potential  energies  when 
it  reaches  the  earth  ;  also  the  work  done  in  the  fall. 

ARTICLE  557. 

Equation  6  should  read  as  follows : 
W  v2       Ox       W  vl   .  Q,  x,  0,  x, 

sm  at  008  at  -- 


78 


ARTICLE    578. 


ARTICLE  578. 

A  method  of  finding  the  moment  of  inertia  of  solids  of  revo- 
lution is  illustrated  in  the  solution  of  the  following  problem.  Find 
the  moment  of  inertia  of  a  cone  whose  altitude  is  h  and  radius 
of  base  r,  about  an  axis  through 
its  vertex  and  perpendicular  to 
its  geometric  axis.  Conceive  a 
portion  of  the  cone  included  be- 
tween two  planes  at  right  angles 
to  its  axis,  and  at  a  distance  tlx 
apart.  Find  the  moment  of  inertia 
of  this  elementary  slice  or  disc,  by 
applying  to  it  the  theorem  of  Art. 
576  A.  M. 

w*y*dx.  x2. 


dx  - 


X 


dx.    This  is  the  formula  for  problem 


If  the  moment  of  inertia  about  the  axis  0  X  were  required, 
the  axis  of  inertia  would  pass  through  the  center  of  gravity  of 
the  disc  and  be  perpendicular  to  its  bases.  We  should  there- 

fore have  Ix  =  —  -  /     y* 
2    u   o 

I  of  this  Article. 

PROBLEM  III.  —  Although  the  ellipsoid  is  not  a  solid  of  revo- 
lution, its  moment  of  inertia  may  be  found  by  the  method  above 
explained. 

Let  the  origin  be  at  the  center  of  the  ellipsoid  and  let  0  X, 
0  Y  and  OZ  be  in  the  direction  of  the  semi-axes  a,  I  and  c  re- 
spectively. A  section  perpendicular  to  O  X  is  an  ellipse  whose 
semi-axes  are  y  and  z.  The  moment  of  inertia  of  the  elemen- 
tary slice  with  respect  to  0  X  is  therefore 

w*y*z   dx  +  W  *y  Z 


d  x. 


W_7T       /»    ° 
TV    -a' 


(See  problem  in  Notes  on  Art.  95.) 
T/_>3^' 


The  values  of  z  and  y  are  taken  from  the  following  equations, 
j  a2  z2  4-  c2  x2  =  a2  c2 
\  «*  if  -I-  J*  x1  =  a2  V  ' 


ARTICLES  579-590.  79 


4  w  TT  a  b  c  (b*  -f-  c2) 

15 

Let  the  student  perform  the  same  problem  by  integrating  the 
following  formula  according  to  the  directions  in  Note  on  Art.  83, 
assigning  the  proper  limits. 

I  =  10     *  dx  dy  dz  (y2  +  z2). 


Theorem.  —  The  actual  energy  of  a  body  revolving  about  a 
fixed  axis  is  found  by  multiplying  its  moment  of  inertia  by  the 
square  of  its  angular  velocity  and  dividing  the  product  by  twice 
the  force  of  gravity. 

Let  p  =  the  radius  of  gyration  of  the  body,  a  its  angular  ve- 
locity, and  W  its  weight.  If  the  mass  of  the  body  were  con- 
centrated at  the  center  of  gyration,  (that  is,  at  a  distance  p  from 
the  axis  of  rotation)  its  velocity  would  be  a  p.  Its  actual  energy 

"W"  r2  r/2 

would  be  J|  a'  P2  =  ---  .  W  p«  =  £_  I. 

If  the  axis  of  rotation  traverses  the  center  of  gravity  of  the 
body  and  is  a  "  line  of  symmetry  of  the  figure  of  the  body,"  the 
values  of  I  and  p  may  be  taken  from  the  columns  headed  I0  and 
p0  in  the  table  of  Art.  578  A.  M.  (See  Art.  589  A.  M.) 

ARTICLE  579. 

V  -4-  c2  W  -f-  c 

Read    —  —  —  instead  of  —  —  -  under  example. 
6  6 

ARTICLE  590. 
Equation  5.  —  From  equation  ('!)  of  Art.  585,  we  haves2  =  -- 

From  equation  (6)  of  Art.  588  we  have  cos  0  =  -  =  -  =—  . 

*       VF+K" 

_          «2  (I2  +  K2) 

~T-     "P 

Or  1  =  —  (P  -f  K2)  •  .-.  -1—  P  +  K2  =  11  cos-  «  -f  I2  cos2 
/3  +  II  cos2  y.  Equation  (2) 

NOTE.—  The  moment  of  inertia  as  here  used  is  defined  in  Arts.  571,  572,  and  573,  A.  M. 


80  ARTICLES  606-648. 

But  I  =  I,  cos2  a  +  I2  cos2  /3  +  I3  cos2  r.   (Eq.  (3),  Art.  585) 

,;.  IL^_-  +k^+lL?£r  =  u  ^  «  +  n  cos*  ,3  +  is 

COS2  y. 


cos2  y  =  0. 

ARTICLE  606. 

First  Equation.  —  The  moment  of  the  deviating  couple  in 
terms  of  E  is  M  =  2  E  tan  a.  The  value  of  the  constant  E  is 
the  sum  of  its  components  with  reference  to  the  axes  C  G  and 

C  E.  and  these  components  are  -  -  -  cos2  a  and  -r  —  -   cos2 

9  \9 

(180  -  (90°  -  «))  =  -  ^-S   sin2  «     .-.   M  ==  —  (I,   -   I2) 

w 

COS  a  sin  a  =   —  a2  (p*  —  p22)  cos  a  sin  a. 

ARTICLE  638. 

Formulae  for  computing  the  discharge  and  diameter  of  pipes, 
are  given  in  Article  450  of  Rankine's  Civil  Engineering. 

ARTICLE  648. 

Equations  (1),  (2),  and  (3).  —  These  equations  may  be  obtained 
as  follows  :  The  velocity  of  the  jet  before  it  meets  the  surface 
is  v.  When  the  surface  has  changed  its  direction  through  an 
angle  /3,  the  component  of  its  velocity  in  the  original  direction 
of  the  jet  is  v  cos  /3.  The  change  of  velocity  is,  therefore,  v  — 
v  cos  ft  =  v  (1  —  cos  /3).  Hence,  the  change  of  momentum 

p  Q 

which  measures  the  force  in  this  direction,  is  F,  =  —  —  i 

9 

(1  —  cos  /3).     The  change  of  velocity  in  a  direction  perpendicu 
lar    the    original   direction   of    the   jet   is  v  sin   j3     .•.  Fy  = 


sin  /3.     The  resultant  F  =   v  Fx2  -j-  F/  = 


<J  g    r. 

PQ 


—  COS  Ij)  =• 


P   V 


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